Thinking Process

(i) First, we use the formula $\sin \theta=\sqrt{1-\cos ^{2} \theta}$ to get the value of $\sin \theta$.

(ii) Second, we use the formula $\tan \theta=\dfrac{\sin \theta}{\cos \theta}$ to get the value of $\tan \theta$.

Solution

(b) Given, $\cos A=\dfrac{4}{5}$

$ \begin{aligned} & \therefore \quad \sin A=\sqrt{1-\cos ^{2} A} \\ & =\sqrt{1-\dfrac{4}5^{2}}=\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5} \\ & \text{ Now, } \tan A=\dfrac{\sin A}{\cos A}=\dfrac{\dfrac{3}{5}}{\dfrac{5}{5}}=\dfrac{3}{4} \\ & \because \sin ^{2} A+\cos ^{2} A=1 \\ & \therefore \sin A=\sqrt{1-\cos ^{2} A} \end{aligned} $

Hence, the required value of $\tan A$ is $3 / 4$.

Thinking Process

(i) First, we use the formula $\cos \theta=\sqrt{1-\sin ^{2} \theta}$ to get the value of $\cos \theta$.

(ii) Now, we use the trigonometric ratio $\cot \theta=\dfrac{\cos \theta}{\sin \theta}$ to get the value of $\cot \theta$.

Solution

(a) Given, $\quad \sin A=\dfrac{1}{2}$

$ \begin{array}{rl} \therefore \cos A = \sqrt{1-\sin ^{2} A}=\sqrt{1-\dfrac{1}2^{2}} \\ & =\sqrt{1-\dfrac{1}{4}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2} \quad[\because \sin ^{2} A+\cos ^{2}=1 \Rightarrow \cos A=\sqrt{1-\sin ^{2} A}] \\ \text{ Now, } \cot A = \dfrac{\cos A}{\sin A}=\dfrac{\dfrac{\sqrt{3}}{\dfrac{1}{2}}}{\dfrac{1}{2}}=\sqrt{3} \end{array} $

Hence, the required value of $\cot A$ is $\sqrt{3}$.

Thinking Process

We see that, the given trigonometric angle of the ratio are the reciprocal in the sense of sign. Then, use the following formulae

(i) $cosec(90^{\circ}-\theta)=\sec \theta$ (ii) $\cot (90^{\circ}-\theta)=\tan \theta$

Solution

(b) Given, expression $=cosec(75^{\circ}+\theta)-\sec (15^{\circ}-\theta)-\tan (55^{\circ}+\theta)+\cot (35^{\circ}-\theta)$

$ \begin{aligned} & =cosec[90^{\circ}-(15^{\circ}-\theta)]-\sec (15^{\circ}-\theta)-\tan (55^{\circ}+\theta)+\cot {90^{\circ}-(55^{\circ}+\theta)} \\ & =\sec (15^{\circ}-\theta)-\sec (15^{\circ}-\theta)-\tan (55^{\circ}+\theta)+\tan (55^{\circ}+\theta) \\ & =0 \quad[\because cosec(90^{\circ}-\theta)=\sec \theta \text{ and } \cot (90^{\circ}-\theta)=\tan \theta] \end{aligned} $

Hence, the required value of the given expression is 0 .

Solution

(c) Given,

$ \begin{array} \sin \theta = \dfrac{a}{b} \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}] \\ \cos \theta = \sqrt{1-\sin ^{2} \theta} \\ =\sqrt{1-(\dfrac{a}{b})^2}=\sqrt{1-\dfrac{a^{2}}{b^{2}}}=\dfrac{\sqrt{b^{2}-a^{2}}}{b} \end{array} $

$ \therefore \quad \cos \theta=\sqrt{1-\sin ^{2} \theta} $

Solution

(b) Given, $\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $[\because \cos 90^{\circ}=0]$

$\Rightarrow$ $\alpha+\beta=90^{\circ}$

$\Rightarrow \quad \alpha=90^{\circ}-\beta$

Now,

$\sin (\alpha-\beta)=\sin (90^{\circ}-\beta-\beta)$

$=\sin (90^{\circ}-2 \beta)$

$=\cos 2 \beta$

$[\because \sin (90^{\circ}-\theta)=\cos \theta]$

Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

Thinking Process

Use the transformation $\tan (90^{\circ}-\theta)=\cot \theta$ from greater than trigonometric angle $\tan 45^{\circ}$ after that we use the trigonometric ratio, $\cot \theta=\dfrac{1}{\tan \theta}$.

Solution

(b) $\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 89^{\circ}$

$ \begin{aligned} &= \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 44^{\circ} \cdot \tan 45^{\circ} \cdot \tan 46^{\circ} \ldots \tan 87^{\circ} \cdot \tan 88^{\circ} \cdot \tan 89^{\circ} \\ &= \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 44^{\circ} \cdot(1) \cdot \tan (90^{\circ}-44^{\circ}) \ldots \tan (90^{\circ}-3^{\circ}) \cdot \\ & \tan (90^{\circ}-2^{\circ}) \cdot \tan (90^{\circ}-1^{\circ}) \quad(\because \tan 45^{\circ}=1) \\ &=\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 44^{\circ} \cdot(1) \cdot \cot 44^{\circ} \ldots \ldots \cot 3^{\circ} \cdot \cot 2^{\circ} \cdot \cot 1^{\circ} \\ &= \quad[\because \tan (90^{\circ}-\theta)=\cot \theta] \\ &= \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 44^{\circ}(1) \cdot \dfrac{1}{\tan 44^{\circ}} \cdots \dfrac{1}{\tan 30^{\circ}} \cdot \dfrac{1}{\tan 2^{\circ}} \cdot \dfrac{1}{\tan 1^{\circ}} \quad \because \cot \theta=\dfrac{1}{\tan \theta} \\ &=1 \end{aligned} $

Solution

(c) Given,

$ \begin{aligned} & \cos 9 \alpha=\sin \alpha \text{ and } 9 \alpha<90^{\circ} \text{ i.e., acute angle. } \\ & \sin (90^{\circ}-9 \alpha)=\sin \alpha \quad[\because \cos A=\sin (90^{\circ}-A)] \\ \end{aligned} $

$ \begin{aligned} \Rightarrow & 90^{\circ}-9 \alpha = \alpha \\ \Rightarrow & 10 \alpha = 90^{\circ} \\ \Rightarrow & \alpha = 9^{\circ} \end{aligned} $

$ \therefore \quad \tan 5 \alpha=\tan (5 \times 9^{\circ})=\tan 45^{\circ}=1 \quad[\because \tan 45^{\circ}=1]$

not in between.

Solution

(a) We know that, in $\triangle A B C$, sum of three angles $=180^{\circ}$ i.e., $\angle A+\angle B+\angle C=180^{\circ}$ But right angled at $C$ i.e., $\angle C=90^{\circ}$ [given] $\angle A+\angle B+90^{\circ}=180^{\circ}$ $\begin{matrix} \Rightarrow & A+B = 90^{\circ} \\ \therefore & \cos (A+B) = \cos 90^{\circ}=0\end{matrix} \quad[\because \angle A=A]$

Solution

(a) Given, $\sin A+\sin ^{2} A=1$

$\Rightarrow \quad \sin A=1-\sin ^{2} A=\cos ^{2} A$

$[\because \sin ^{2} \theta+\cos ^{2} \theta=1]$

On squaring both sides, we get

$ \begin{matrix} \Rightarrow & 1-\cos ^{2} A=\cos ^{4} A \\ \Rightarrow & \cos ^{2} A+\cos ^{4} A=1 \end{matrix} $

Solution

(d) Given,

$ \sin \alpha=\dfrac{1}{2}=\sin 30^{\circ} $

$ \begin{aligned} & \Rightarrow \quad \alpha=30^{\circ} \\ & \text{ and } \quad \cos \beta=\dfrac{1}{2}=\cos 60^{\circ} \\ & \Rightarrow \quad \beta=60^{\circ} \\ & \therefore \quad \alpha+\beta=30^{\circ}+60^{\circ}=90^{\circ} \end{aligned} $

Solution

(b) Given expression, $\dfrac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$

$ \begin{aligned} & =\dfrac{\sin ^{2} 22^{\circ}+\sin ^{2}(90^{\circ}-22^{\circ})}{\cos ^{2}(90^{\circ}-68^{\circ})+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin (90^{\circ}-63^{\circ}) \\ & =\dfrac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ} \quad \begin{matrix} [\because \sin (90^{\circ}-\theta)=\cos \theta \text{ and } \cos (90^{\circ}-\theta)=\sin \theta] \end{matrix} \\ & =\dfrac{1}{1}+(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}) \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ & =1+1=2 \end{aligned} $

Solution

(c) Given,

$4 \tan \theta=3$

$\Rightarrow \quad \tan \theta=\dfrac{3}{4}$

$\therefore \quad \dfrac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\dfrac{4 \dfrac{\sin \theta}{\cos \theta}-1}{4 \dfrac{\sin \theta}{\cos \theta}+1}$

[divide by $\cos \theta$ in both numerator and denominator]

$=\dfrac{4 \tan \theta-1}{4 \tan \theta+1} \quad [\because \tan \theta=\dfrac{\sin \theta}{\cos \theta}]$

$=\dfrac{4 \dfrac{3}{4}-1}{4 \dfrac{3}{4}+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2} \quad$

[put the value from Eq. (i)]

Thinking Process

Firstly, from $\sin \theta-\cos \theta=0$ get the value of $\theta$. After that put the value of $\theta$ in the given expression to get the desired result.

Solution

(c) Given,

$\Rightarrow \quad \tan \theta=1 \quad \because \tan \theta=\dfrac{\sin \theta}{\cos \theta}$ and $\tan 45^{\circ}=1$

$\Rightarrow \quad \tan \theta=\tan 45^{\circ}$

$\therefore \quad \theta=45^{\circ}$

Now,

$\sin \theta-\cos \theta=0$

$\sin ^{4} \theta+\cos ^{4} \theta=\sin ^{4} 45^{\circ}+\cos ^{4} 45^{\circ}$

$ \begin{matrix} =\dfrac{1}{\sqrt{2}}^{4}+\dfrac{1}{\sqrt{2}}^{4} \quad \because \sin 45^{\circ}=\cos 45^{\circ}=\dfrac{1}{\sqrt{2}} \\ =\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2} \end{matrix} $

Solution

(b) $\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)=\cos [90^{\circ}-(45^{\circ}+\theta)]-\cos (45^{\circ}-\theta) \\ [\because \cos (90^{\circ}-\theta)=\sin \theta]$ $=\cos (45^{\circ}-\theta)-\cos (45^{\circ}-\theta)$ $=0$

Solution

(a) Let $B C=6 m$ be the height of the pole and $A B=2 \sqrt{3} m$ be the length of the shadow on the ground. let the Sun’s makes an angle $\theta$ on the ground.

Now, in $\triangle B A C$,

$\tan \theta=\dfrac{B C}{A B}$

$\Rightarrow \quad \tan \theta=\dfrac{6}{2 \sqrt{3}}=\dfrac{3}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow \quad \tan \theta=\dfrac{3 \sqrt{3}}{3}=\sqrt{3}=\tan 60^{\circ}$

$\therefore \quad \theta=60^{\circ}$

$[\because \tan 60^{\circ}=\sqrt{3}]$

Hence, the Sun’s elevation is $60^{\circ}$.

Solution

True

$ \dfrac{\tan 47^{\circ}}{\cot 43^{\circ}}=\dfrac{\tan (90^{\circ}-43^{\circ})}{\cot 43^{\circ}}=\dfrac{\cot 43^{\circ}}{\cot 43^{\circ}}=1 \quad[\because \tan (90^{\circ}-\theta)=\cot \theta] $

Solution

False

$ \begin{array}{lr} \cos ^{2} 23^{\circ}-\sin ^{2} 67^{\circ} = (\cos 23^{\circ}-\sin 67^{\circ})(\cos 23^{\circ}+\sin 67^{\circ}) \quad[\because(a^{2}-b^{2})=(a-b)(a+b)] \\ =[\cos 23^{\circ}-\sin (90^{\circ}-23^{\circ})](\cos 23^{\circ}+\sin 67^{\circ}) \\ =(\cos 23^{\circ}-\cos 23^{\circ})(\cos 23^{\circ}+\sin 67^{\circ}) \quad[\because \sin (90^{\circ}-\theta)=\cos \theta] \\ =0 \cdot(\cos 23^{\circ}+\sin 67^{\circ})=0 \end{array} $

which may be either positive or negative.

Solution

False

We know that, $\sin \theta$ is increasing when, $0^{\circ} \leq \theta \leq 90^{\circ}$ and $\cos \theta$ is decreasing when, $0^{\circ} \leq \theta \leq 90^{\circ}$.

$\therefore \quad \sin 80^{\circ}-\cos 80^{\circ}>0$

[positive]

Solution

True

$ \begin{aligned} & \sqrt{(1-\cos ^{2} \theta) \sec ^{2} \theta}=\sqrt{\sin ^{2} \theta \cdot \sec ^{2} \theta} \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ & =\sqrt{\sin ^{2} \theta \cdot \dfrac{1}{\cos ^{2} \theta}}=\sqrt{\tan ^{2} \theta}=\tan \theta \quad \because \sec \theta=\dfrac{1}{\cos \theta}, \tan \theta=\dfrac{\sin \theta}{\cos \theta} \end{aligned} $

Solution

True

$\begin{array}{lrlrl} \because & \cos A+\cos ^2 A = 1 & \\ \Rightarrow & \cos A=1-\cos ^2 A = \sin ^2 A & & \left[\sin ^2 A+\cos ^2 A=1\right]\\ \Rightarrow & \cos ^2 A = \sin ^4 A & \\ \Rightarrow & 1-\sin ^2 A = \sin ^4 A & & \\ \Rightarrow & \sin ^2 A+\sin ^4 A = 1 & & \left[\cos ^2 A=1-\sin ^2 A\right]\end{array}$

Solution

False

$ \begin{array}{ll} \text{ LHS } = (\tan \theta+2)(2 \tan \theta+1) \\ =2 \tan ^{2} \theta+4 \tan \theta+\tan \theta+2 \\ =2(\sec ^{2} \theta-1)+5 \tan \theta+2 & [\because \sec ^{2} \theta-\tan ^{2} \theta=1]\\ =2 \sec ^{2} \theta+5 \tan \theta=\text{ RHS } \end{array} \quad $

Solution

False

To understand the fact of this question, consider the following example

I. A tower $2 \sqrt{3} m$ high casts a shadow $2 m$ long on the ground, then the Sun’s elevation is $60^{\circ}$.

$ \begin{matrix} \text{ In } \triangle A C B, & & \tan \theta = \dfrac{A B}{B C}=\dfrac{2 \sqrt{3}}{2} \\ \Rightarrow & \tan \theta = \sqrt{3}=\tan 60^{\circ} \\ \therefore & & \theta = 60^{\circ} \end{matrix} $

II. A same hight of tower casts a shadow $4 m$ more from preceding

$ \begin{aligned} & \text{ In } \triangle A P B, \quad \tan \theta=\dfrac{A B}{P B}=\dfrac{A B}{P C+C B} \\ & \Rightarrow \quad \tan \theta=\dfrac{2 \sqrt{3}}{4+2}=\dfrac{2 \sqrt{3}}{6} \\ & \Rightarrow \quad \tan \theta=\dfrac{\sqrt{3}}{3} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{3}{3 \sqrt{3}} \\ & \Rightarrow \quad \tan \theta=\dfrac{1}{\sqrt{3}}=\tan 30^{\circ} \\ & \therefore \quad \theta=30 \end{aligned} $

Hence, we conclude from above two examples that if the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is decreasing.

Alternate Method

False, we know that, if the elevation moves towards the tower, it increases and if its elevation moves away the tower, it decreases. Hence, if the shadow of a tower is increasing, then the angle of elevation of a Sun is not increasing.

Solution

False

From figure, we observe that, a man standing on a platform at point $P, 3 m$ above the surface of a lake observes a cloud at point $C$. Let the height of the cloud from the surface of the platform is $h$ and angle of elevation of the cloud is $\theta_1$.

Now at same point $P$ a man observes a cloud reflection in the lake at this time the height of reflection of cloud in lake is $(h+3)$ because in lake platform height is also added to reflection of cloud.

So, angle of depression is different in the lake from the angle of elevation of the cloud above the surface of a lake.

$ \begin{matrix} \text{ In } \triangle M P C, & \tan \theta_1=\dfrac{C M}{P M}=\dfrac{h}{P M} \\ \Rightarrow & \dfrac{\tan \theta_1}{h}=\dfrac{1}{P M} & \ldots \text{(ii)}\\ \text{ In } \triangle C P M, & \tan \theta_2=\dfrac{C M}{P M}=\dfrac{O C}{F} \\ \Rightarrow & \dfrac{\tan \theta_2}{h+3}=\dfrac{1}{P M} \end{matrix} $

Hence,

Alternate Method

$ \dfrac{\tan \theta_1}{h}=\dfrac{\tan \theta_2}{h+3} $

False, we know that, if $P$ is a point above the lake at a distance $d$, then the reflection of the point in the lake would be at the same distance $d$. Also, the angle of elevation and depression from the surface of the lake is same.

Here, the man is standing on a platform $3 m$ above the surface, so its angle of elevation to the cloud and angle of depression to the reflection of the cloud is not same.

Thinking Process

Use the relation between arithmetic mean and geometric mean i.e., AM > GM.

Solution

False

Given, $a$ is a positive number and $a \neq 1$, then $A M>G M$

$ \Rightarrow \quad \dfrac{a+\dfrac{1}{a}}{2}>\sqrt{a \cdot \dfrac{1}{a}} \Rightarrow a+\dfrac{1}{a}>2 $

[since, AM and GM of two number’s $a$ and $b$ are $\dfrac{(a+b)}{2}$ and $\sqrt{a b}$, respectively]

$ \because 2 \sin \theta=a+\dfrac{1}{a} $

$ \begin{matrix} \Rightarrow & 2 \sin \theta>2 \\ \Rightarrow & \sin \theta>1 \end{matrix} $

which is not possible.

$ [\because-1 \leq \sin \theta \leq 1] $

Hence, the value of $2 \sin \theta$ can not be $a+\dfrac{1}{a}$.

Solution

False

Given, $a$ and $b$ are two distinct numbers such that $a b>0$.

Using,

$AM>GM$

[since, AM and GM of two number $a$ and $b$ are $\dfrac{a+b}{2}$ and $\sqrt{a b}$, respectively]

$\Rightarrow \quad \dfrac{a^{2}+b^{2}}{2}>\sqrt{a^{2} \cdot b^{2}}$

$\Rightarrow \quad a^{2}+b^{2}>2 a b$

$\Rightarrow \quad \dfrac{a^{2}+b^{2}}{2 a b}>1 \quad \because \cos \theta=\dfrac{a^{2}+b^{2}}{2 a b}$

$\Rightarrow \quad \cos \theta>1 \quad[\because-1 \leq \cos \theta \leq 1]$

which is not possible.

Hence, $\quad \cos \theta \neq \dfrac{a^{2}+b^{2}}{2 a b}$

Thinking Process

(i) Firstly, we find relation between $h$ and $x$ by putting angle of elevation is 30 in case 1.

(ii) After that we take double height and taking angle of elevation is $\theta$. Now, use the relation (i) and get the desired result.

Solution

False

Case I Let the height of the tower is $h$ and $B C=x m$ In $\triangle A B C$,

$ \begin{aligned} & \tan 30^{\circ}=\dfrac{A C}{B C}=\dfrac{h}{x} \\ & \Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{h}{x} \end{aligned} $

Case II By condition, the height of the tower is doubled. i.e., $P R=2 h$.

$ \begin{matrix} \text{ In } \triangle P Q R, & & \tan \theta = \dfrac{P R}{Q R}=\dfrac{2 h}{x} \\ \Rightarrow & & \tan \theta = \dfrac{2}{x} \times \dfrac{x}{\sqrt{3}} \\ \Rightarrow & & \tan \theta = \dfrac{2}{\sqrt{3}}=1.15 \\ & \therefore & \theta = \tan ^{-1}(1.15)<60^{\circ} \end{matrix} $

$ \Rightarrow \quad \tan \theta=\dfrac{2}{x} \times \dfrac{x}{\sqrt{3}} \quad \because h=\dfrac{x}{\sqrt{3}} \text{, from Eq. (i) } $

Hence, the required angle is not doubled.

Solution

True

Case I Let the height of a tower be $h$ and the distance of the point of observation from its foot is $x$.

In $\triangle A B C$,

$ \begin{aligned}\tan \theta_1 = \dfrac{A C}{B C}=\dfrac{h}{x} \\ \Rightarrow \quad \theta_1 = \tan ^{-1} \dfrac{h}{x} & \ldots \text{(i)} \end{aligned} $

Case II Now, the height of a tower increased by $10 %=h+10 %$ of $h=h+h \times \dfrac{10}{100}=\dfrac{11 h}{10}$ and the distance of the point of observation from its foot $=x+10 %$ of $x$

$ \begin{matrix}& =x+x \times \dfrac{10}{100}=\dfrac{11 x}{10} \\ \ln \triangle P Q R, \quad \tan \theta_2 = \dfrac{P R}{Q R}=\dfrac{\dfrac{11 h}{10}}{\dfrac{11 x}{10}} \\ \Rightarrow \quad & \tan \theta_2 = \dfrac{h}{x} \\ \Rightarrow \quad & \theta_2 = \tan ^{-1} \dfrac{h}{x} & \ldots \text{(ii)} \end{matrix} $

From Eqs. (i) and (ii),

$ \theta_1=\theta_2 $

Hence, the required angle of elevation of its top remains unchanged.

Solution

$LHS=\dfrac{\sin \theta}{1+\cos \theta}+\dfrac{1+\cos \theta}{\sin \theta}=\dfrac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{\sin \theta(1+\cos \theta)}$

$ \begin{matrix} =\dfrac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{\sin \theta(1+\cos \theta)} & {[\because(a+b)^{2}=a^{2}+b^{2}+2 a b]} \\ =\dfrac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)} & \\ =\dfrac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}=\dfrac{2}{\sin \theta} & \\ =2 cosec \theta=\text{ RHS } & \end{matrix} $

Solution

$L H S=\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=\dfrac{\tan A(1-\sec A-1-\sec A)}{(1+\sec A)(1-\sec A)}$

$ \begin{aligned} & =\dfrac{\tan A(-2 \sec A)}{(1-\sec ^{2} A)}=\dfrac{2 \tan A \cdot \sec A}{(\sec ^{2} A-1)} \\ & .(a+b)(a-b)=a^{2}-b^{2}] \\ & =\dfrac{2 \tan A \cdot \sec A}{\tan ^{2} A} \quad[\because \sec ^{2} A-\tan ^{2} A=1] \quad \because \sec \theta=\dfrac{1}{\cos \theta} \text{ and } \tan \theta=\dfrac{\sin \theta}{\cos \theta} \\ & =\dfrac{2 \sec A}{\tan A}=\dfrac{2}{\sin A}=2 cosec A=\text{ RHS } \quad \because cosec \theta=\dfrac{1}{\sin \theta} \end{aligned} $

Thinking Process

We know that, $\tan \theta=\dfrac{\text{ Perpendicular }}{\text{ Base }}$. Now, using Pythagoras theorem, get the value of hypotenuse. i.e., $(\text{ Hypotenuse })^{2}=(\text{ Base })^{2}+(\text{ Perpendicular })^{2}$ and then find the value of trigonometric ratios $\sin \theta$ and $\cos \theta$ and get the desired result.

Solution

Given,

$ \tan A=\dfrac{3}{4}=\dfrac{P}{B}=\dfrac{\text{ Perpendicular }}{\text{ Base }} $

Let $\quad P=3 k$ and $B=4 k$

By Pythagoras theorem,

$ \begin{aligned} H^{2} = P^{2}+B^{2}=(3 k)^{2}+(4 k)^{2} \\ & =9 k^{2}+16 k^{2}=25 k^{2} \\ \Rightarrow \quad H = 5 k \quad \text{ [since, side cannot be negative] } \\ \therefore \quad \sin A = \dfrac{P}{H}=\dfrac{3 k}{5 k}=\dfrac{3}{5} \quad \text{ and } \cos A=\dfrac{B}{H}=\dfrac{4 k}{5 k}=\dfrac{4}{5} \\ \text{ Now, } \quad \sin A \cos A = \dfrac{3}{5} \cdot \dfrac{4}{5}=\dfrac{12}{25} \end{aligned} $

Hence proved.

Solution

$LHS=(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)$

$ \begin{matrix} =(\sin \alpha+\cos \alpha) \dfrac{\sin \alpha}{\cos \alpha}+\dfrac{\cos \alpha}{\sin \alpha} & \because \tan \theta=\dfrac{\sin \theta}{\cos \theta} \text{ and } \cot \theta=\dfrac{\cos \theta}{\sin \theta} \\ =(\sin \alpha+\cos \alpha) \dfrac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin \alpha \cdot \cos \alpha} & \\ =(\sin \alpha+\cos \alpha) \cdot \dfrac{1}{(\sin \alpha \cdot \cos \alpha)} \\ {[\because \sin ^{2} \theta+\cos ^{2} \theta=1]} & \\ =\dfrac{1}{\cos \alpha}+\dfrac{1}{\sin \alpha} & \\ =\sec \alpha+\cos \alpha=\text{ RHS } & \end{matrix} $

Solution

RHS $=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}=(\sqrt{3})^{3}-2 \dfrac{\sqrt{3}}{2}=3 \sqrt{3}-\sqrt{3}=2 \sqrt{3}$

LHS $.=(\sqrt{3}+1)(3-\cot 30^{\circ})=(\sqrt{3} + 1)(3 - \sqrt{3})$

${[\because \tan 60^{\circ}}=\sqrt{3} \text{ and } \sin 60^{\circ}=\dfrac{\sqrt{3}}{2}]$

$=(\sqrt{3}+1) \sqrt{3}(\sqrt{3}-1) \cot 30^{\circ}=\sqrt{3}$

$=\sqrt{3}(\sqrt{3})^{2}-1)=\sqrt{3}(3-1)=2 \sqrt{3}$

$ \therefore \quad \text{ LHS }=\text{ RHS } $

Hence proved.

Solution

LHS $=1+\dfrac{\cot ^{2} \alpha}{1+cosec \alpha}=1+\dfrac{\cos ^{2} \alpha / \sin ^{2} \alpha}{1+1 / \sin \alpha} \quad \because \cot \theta=\dfrac{\cos \theta}{\sin \theta}$ and $cosec \theta=\dfrac{1}{\sin \theta}$

$ \begin{matrix} =1+\dfrac{\cos ^{2} \alpha}{\sin \alpha(1+\sin \alpha)}=\dfrac{\sin \alpha(1+\sin \alpha)+\cos ^{2} \alpha}{\sin \alpha(1+\sin \alpha)} & \\ =\dfrac{\sin \alpha+(\sin ^{2} \alpha+\cos ^{2} \alpha)}{\sin \alpha(1+\sin \alpha)} & {[\because \sin ^{2} \theta+\cos ^{2} \theta=1]} \\ =\dfrac{(\sin \alpha+1)}{\sin \alpha(\sin \alpha+1)}=\dfrac{1}{\sin \alpha} & \because cosec \theta=\dfrac{1}{\sin \theta} \\ =cosec \alpha=\text{ RHS } & \end{matrix} $

Solution

$LHS=\tan \theta+\tan (90^{\circ}-\theta)$

$ \begin{matrix} =\tan \theta+\cot \theta=\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} & \\ =\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} & \because \tan \theta=\dfrac{\sin \theta}{\cos \theta} \text{ and } \cot \theta=\dfrac{\cos \theta}{\sin \theta} \\ =\dfrac{1}{\sin \theta \cos \theta} & {[\because \sec \theta=\dfrac{1}{\cos \theta} \text{ and } \cos \theta=\dfrac{1}{\sin \theta}.} \\ =\sec \theta cosec \theta & {[\because \sec (90^{\circ}-\theta)=cosec \theta]} \\ =\sec \theta \sec (90^{\circ}-\theta)=\text{ RHS } & .\cos ^{2} \theta=1] \\ & \end{matrix} $

Solution

Let the angle of elevation of the Sun is $\theta$.

Given, $\quad$ height of pole $=h$

Now, in $\triangle A B C$,

$ \begin{matrix} \text{ Now, in } \triangle A B C, \\ & \tan \theta=\dfrac{A B}{B C}=\dfrac{h}{\sqrt{3} h} \\ \Rightarrow \quad \tan \theta=\dfrac{1}{\sqrt{3}}=\tan 30^{\circ} \Rightarrow \theta=30^{\circ} \end{matrix} $

Hence, the angle of elevation of the Sun is $30^{\circ}$.

Thinking Process

From the given equation, get the value of $\theta$ and put the value of $\theta$ in the given expression, we get the required value.

Solution

Given that,

$\sqrt{3} \tan \theta=1$

$\Rightarrow$

$ \tan \theta=\dfrac{1}{\sqrt{3}}=\tan 30^{\circ} $

$\Rightarrow$

$ \theta=30^{\circ} $

Now,

$ \sin ^{2} \theta-\cos ^{2} \theta=\sin ^{2} 30^{\circ}-\cos ^{2} 30^{\circ} $

$ \begin{aligned} & =(\dfrac{1}{2})^2 - (\dfrac{\sqrt{3}}{2})^2 \\ & =\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{1-3}{4}=-\dfrac{2}{4}=-\dfrac{1}{2} \end{aligned} $

Solution

Given that, the height of the ladder $=15 m$

Let the height of the vertical wall $=h$

and the ladder makes an angle of elevation $60^{\circ}$ with the wall i.e., $\theta=60^{\circ}$ In $\triangle Q P R$,

$ \cos 60^{\circ}=\dfrac{P R}{P Q}=\dfrac{h}{15} $

$ \begin{matrix} \Rightarrow & \dfrac{1}{2}=\dfrac{h}{15} \\ \Rightarrow & h=\dfrac{15}{2} m . \end{matrix} $

Hence, the required height of the wall $\dfrac{15}{2} m$

Solution

$(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)=(1+\tan ^{2} \theta)(1-\sin ^{2} \theta)$

$ \begin{aligned} & =\sec ^{2} \theta \cdot \cos ^{2} \theta \\ & \quad[\because 1+\tan ^{2} \theta=\sec ^{2} \theta \text{ and } \cos ^{2} \theta+\sin ^{2} \theta=1] \\ & =\dfrac{1}{\cos ^{2} \theta} \cdot \cos ^{2} \theta=1 \quad \because \sec \theta=\dfrac{1}{\cos \theta} \end{aligned} $

Solution

Given,

$ 2 \sin ^{2} \theta-\cos ^{2} \theta=2 $

$\Rightarrow \quad 2 \sin ^{2} \theta-(1-\sin ^{2} \theta)=2$

$[\because \sin ^{2} \theta+\cos ^{2} \theta=1]$

$\Rightarrow \quad 2 \sin ^{2} \theta+\sin ^{2} \theta-1=2$

$\Rightarrow \quad 3 \sin ^{2} \theta=3$

$\Rightarrow \quad \sin ^{2} \theta=1 \quad[\because \sin 90^{\circ}=1]$

$\Rightarrow \quad \sin \theta=1=\sin 90^{\circ}$

$\therefore \quad \theta=90^{\circ}$

Solution

LHS $=\dfrac{\cos ^{2}(45^{\circ}+\theta)+\cos ^{2}(45^{\circ}-\theta)}{\tan (60^{\circ}+\theta) \cdot \tan (30^{\circ}-\theta)}$

$ \begin{aligned} & =\dfrac{\cos ^{2}(45^{\circ}+\theta)+[\sin {90^{\circ}-(45^{\circ}-\theta)}]^{2}}{\tan (60^{\circ}+\theta) \cdot \cot {90^{\circ}-(30^{\circ}-\theta)}} \\ & =\dfrac{\cos ^{2}(45^{\circ}+\theta)+\sin ^{2}(45^{\circ}+\theta)}{\tan (60^{\circ}+\theta) \cdot \cot (60^{\circ}+\theta)} \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ & =\dfrac{1}{\tan (60^{\circ}+\theta) \cdot \dfrac{1}{\tan (60^{\circ}+\theta)}}=1=RHS \quad[\because \cot \theta=1 / \tan \theta] \end{aligned} $

Solution

Let the angle of elevation of the top of the tower from the eye of the observe is $\theta$

Given that,

and

$\Rightarrow$

Now, in $\triangle A P M$,

$\Rightarrow$

$ \therefore \quad \theta=45^{\circ} $

Hence, required angle of elevation of the top of the tower from the eye of the observer is $45^{\circ}$.

Solution

LHS $=\tan ^{4} \theta+\tan ^{2} \theta=\tan ^{2} \theta(\tan ^{2} \theta+1)$

$ \begin{matrix} =\tan ^{2} \theta \cdot \sec ^{2} \theta & {[\because \sec ^{2} \theta=\tan ^{2} \theta+1]} \\ =(\sec ^{2} \theta-1) \cdot \sec ^{2} \theta & {[\because \tan ^{2} \theta=\sec ^{2} \theta-1]} \\ =\sec ^{4} \theta-\sec ^{2} \theta=\text{ RHS } & \end{matrix} $

Thinking Process

Reduce the given equation into $\sin \theta$ and $\cos \theta$ and simplify it. In simplification form, we use the componendo and dividendo rule to get the desired result.

Solution

Given,

$ cosec \theta+\cot \theta=p $

$ \begin{aligned} & \Rightarrow \quad \dfrac{1}{\sin \theta}+\dfrac{\cos \theta}{\sin \theta}=p \quad \because cosec \theta=\dfrac{1}{\sin \theta} \text{ and } \cot \theta=\dfrac{\cos \theta}{\sin \theta} \\ & \Rightarrow \quad \dfrac{1+\cos \theta}{\sin \theta}=\dfrac{p}{1} \\ & \Rightarrow \quad \dfrac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}=\dfrac{p^{2}}{1} \quad \text{ [take square on both sides] } \\ & \Rightarrow \quad \dfrac{1+\cos ^{2} \theta+2 \cos \theta}{\sin ^{2} \theta}=\dfrac{p^{2}}{1} \end{aligned} $

Using componendo and dividendo rule, we get

$ \begin{aligned} & \dfrac{(1+\cos ^{2} \theta+2 \cos \theta)-\sin ^{2} \theta}{(1+\cos ^{2} \theta+2 \cos \theta)+\sin ^{2} \theta} = \dfrac{p^{2}-1}{p^{2}+1} \\ \Rightarrow & \dfrac{1+\cos ^{2} \theta+2 \cos \theta-(1-\cos ^{2} \theta)}{1+2 \cos \theta+(\cos ^{2} \theta+\sin ^{2} \theta)} = \dfrac{p^{2}-1}{p^{2}+1} \\ \Rightarrow & \dfrac{2 \cos ^{2} \theta+2 \cos \theta}{2+2 \cos \theta} = \dfrac{p^{2}-1}{p^{2}+1} \\ \Rightarrow & \dfrac{2 \cos \theta(\cos \theta+1)}{2(\cos \theta+1)} = \dfrac{p^{2}-1}{p^{2}+1} \end{aligned} $

$ \therefore \quad \cos \theta=\dfrac{p^{2}-1}{p^{2}+1} $

Hence proved.

Solution

LHS $=\sqrt{\sec ^{2} \theta+cosec^{2} \theta}$

$ \begin{matrix} =\sqrt{\dfrac{1}{\cos ^{2} \theta}+\dfrac{1}{\sin ^{2} \theta}} & \because \sec \theta=\dfrac{1}{\cos \theta} \text{ and } cosec \theta=\dfrac{1}{\sin \theta} \\ =\sqrt{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}=\sqrt{\dfrac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}} & \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ =\dfrac{1}{\sin \theta \cdot \cos \theta}=\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} & \quad[\because 1=\sin ^{2} \theta+\cos ^{2} \theta] \\ =\dfrac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta}+\dfrac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} & \because \tan \theta=\dfrac{\sin \theta}{\cos \theta} \text{ and } \cot \theta=\dfrac{\cos \theta}{\sin \theta} \\ =\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} & \\ =\tan \theta+\cot \theta=\text{ RHS } \end{matrix} $

Thinking Process

(i) First, we drawn a right angle triangle from the given information in the question.

(ii) Now, apply the suitable trigonometric ratio corresponding to given sides in the triangle and get the required value of which we want.

Solution

Let the height of the tower be $h$.

also,

Given that,

and

Now, in $\triangle P S R$,

$ \begin{matrix} \Rightarrow & & \tan \theta = \dfrac{P R}{S R}=\dfrac{h}{x} \\ \Rightarrow & & \tan \theta = \dfrac{h}{x} \\ \Rightarrow & & x=\dfrac{h}{\tan \theta} & \ldots \text{(i)} \end{matrix} $

Now, in $\triangle P Q R$,

$ \begin{matrix} & \tan 30^{\circ}=\dfrac{P R}{Q R}=\dfrac{P R}{Q S+S R} \\ \Rightarrow & \tan 30^{\circ}=\dfrac{h}{20+x} \\ \Rightarrow & 20+x=\dfrac{h}{\tan 30^{\circ}}=\dfrac{h}{1 / \sqrt{3}} \\ \Rightarrow & 20+x=h \sqrt{3} \end{matrix} $

$ \Rightarrow \quad 20+\dfrac{h}{\tan \theta}=h \sqrt{3} \quad \ldots \text{ (ii) [from Eq. (i)] } $

Since, after moving $20 m$ towards the tower the angle of elevation of the top increases by $15^{\circ}$.

i.e., $\quad \angle P S R=\theta=\angle P Q R+15^{\circ}$

$\Rightarrow \quad \theta=30^{\circ}+15=45^{\circ}$

$\therefore \quad 20+\dfrac{h}{\tan 45^{\circ}}=h \sqrt{3}$

$\Rightarrow \quad 20+\dfrac{h}{1}=h \sqrt{3}$

$\Rightarrow \quad 20=h \sqrt{3}-h$

$\Rightarrow \quad h(\sqrt{3}-1)=20$

$\therefore \quad h=\dfrac{20}{\sqrt{3}-1} \cdot \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

[by rationalisation]

$\Rightarrow \quad=\dfrac{20(\sqrt{3}+1)}{3-1}=\dfrac{20(\sqrt{3}+1)}{2}$

$\Rightarrow \quad=10(\sqrt{3}+1) m$

[from Eq. (i)]

Hence, the required height of tower is $10(\sqrt{3}+1) m$.

Thinking Process

(i) First we reduce the given equation, with the help of trigonometric ratio and identities in the form of $\cot \theta$.

(ii) Now, factorise the quadratic equation in $\cot \theta$ by splitting the middle term and get the desired result.

Solution

Given,

$ 1+\sin ^{2} \theta=3 \sin \theta \cdot \cos \theta $

On dividing by $\sin ^{2} \theta$ on both sides, we get

$ \begin{aligned} & \dfrac{1}{\sin ^{2} \theta}+1=3 \cdot \cot \theta \\ & \because \cot \theta=\dfrac{\cos \theta}{\sin \theta} \\ & \Rightarrow \quad cosec^{2} \theta+1=3 \cdot \cot \theta \\ & cosec \theta=\dfrac{1}{\sin \theta} \\ & \Rightarrow \quad 1+\cot ^{2} \theta+1=3 \cdot \cot \theta \quad[\because cosec^{2} \theta-\cot ^{2} \theta=1] \\ & \Rightarrow \quad \cot ^{2} \theta-3 \cot \theta+2=0 \\ & \Rightarrow \quad \cot ^{2} \theta-2 \cot \theta-\cot \theta+2=0 \quad \text{ [by splitting the middle term] } \\ & \Rightarrow \cot \theta(\cot \theta-2)-1(\cot \theta-2)=0 \\ & \Rightarrow \quad(\cot \theta-2)(\cot \theta-1)=0 \Rightarrow \cot \theta=1 \text{ or } 2 \\ & \Rightarrow \quad \tan \theta=1 \text{ or } \dfrac{1}{2} \\ & \because \tan \theta=\dfrac{1}{\cot \theta} \end{aligned} $

Hence proved.

Thinking Process

Squaring both sides the given equation and use the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$ to change $\sin ^{2} \theta$ into $\cos ^{2} \theta$ and vice-versa. Finally use the identity $(a-b)^{2}=a^{2}+b^{2}-2 a b$ and get the desired result.

Solution

Given,

$ \sin \theta+2 \cos \theta=1 $

On squaring both sides, we get

$ \begin{aligned} & (\sin \theta+2 \cos \theta)^{2}=1 \\ & \Rightarrow \quad \sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cdot \cos \theta=1 \\ & \Rightarrow(1-\cos ^{2} \theta)+4(1-\sin ^{2} \theta)+4 \sin \theta \cdot \cos \theta=1 \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ & \Rightarrow \quad-\cos ^{2} \theta-4 \sin ^{2} \theta+4 \sin \theta \cdot \cos \theta=-4 \\ & \Rightarrow \quad 4 \sin ^{2} \theta+\cos ^{2} \theta-4 \sin \theta \cdot \cos \theta=4 \\ & \Rightarrow \quad(2 \sin \theta-\cos \theta)^{2}=4 \quad[\because a^{2}+b^{2}-2 a b=(a-b)^{2}] \\ & \Rightarrow \quad 2 \sin \theta-\cos \theta=2 \quad \text{ Hence proved. } \end{aligned} $

Thinking Process

Use the concept of complementary angles in the angle of elevation. i.e., if two anlges are complementary to each other, then the sum of both angles is equal to $90^{\circ}$.

$ \begin{matrix} \Rightarrow & \alpha+\beta=90^{\circ} \\ \text{ where, } & \alpha=\theta \text{ and } \beta=(90^{\circ}-\theta) \end{matrix} $

Solution

Let the height of the tower is $h$. and

$\angle A B C=\theta$

Given that,

$B C=s, P C=t$

and angle of elevation on both positions are complementary.

i.e.,

$\angle A P C=90^{\circ}-\theta$

[if two angles are complementary to each other, then the sum of both angles is equal to

Now in $\triangle A B C$,

$ \tan \theta=\dfrac{A C}{B C}=\dfrac{h}{S} $

and in $\triangle A P C$

$ \begin{aligned} & \tan (90^{\circ}-\theta)=\dfrac{A C}{P C} \\ & \cot \theta=\dfrac{h}{t} \\ & \dfrac{1}{\tan \theta}=\dfrac{h}{t} \\ & {[\because \tan (90^{\circ}-\theta)=\cot \theta]} \\ & \because \cot \theta=\dfrac{1}{\tan \theta} \ldots \text{ (ii) } \end{aligned} $

On, multiplying Eqs. (i) and (ii), we get

$ \begin{aligned} \Rightarrow & \tan \theta \cdot \dfrac{1}{\tan \theta} = \dfrac{h}{s} \cdot \dfrac{h}{t} \\ \Rightarrow & \dfrac{h^{2}}{s t} = 1 \\ \Rightarrow & h = s t \\ & h = \sqrt{s t} \end{aligned} $

So, the required height of the tower is $\sqrt{s t}$.

Hence proved.

Thinking Process

Consider the shadow of the tower be $x m$ when the angle of elevation in that position is $60^{\circ}$, when angle of elevation is $30^{\circ}$, then the distance becomes $(50+x) m$. Now, apply suitable trigonometric ratios in the triangle and get the desired result.

Solution

Let the height of the tower be $h$ and $R Q=x m$

Given that,

and

Now, in $\triangle S R Q$,

$\Rightarrow$

$ P R=50 m $

$ \begin{aligned} \angle S P Q = 30^{\circ}, \angle S R Q=60^{\circ} \\ \tan 60^{\circ} = \dfrac{S Q}{R Q} \end{aligned} $

$$ \begin{equation*} \sqrt{3}=\dfrac{h}{x} \Rightarrow x=\dfrac{h}{\sqrt{3}} \tag{i} \end{equation*} $$

and in $\triangle S P Q$,

$\tan 30^{\circ}=\dfrac{S Q}{P Q}=\dfrac{S Q}{P R+R Q}=\dfrac{h}{50+x}$

$\Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{h}{50+x}$

$\Rightarrow \quad \sqrt{3} \cdot h=50+x$

$\Rightarrow \quad \sqrt{3} \cdot h=50+\dfrac{h}{\sqrt{3}}$

[from Eq. (i)]

$\Rightarrow \quad(\sqrt{3}-\dfrac{1}{\sqrt{3}}) h=50$

$\Rightarrow \quad \dfrac{(3-1)}{\sqrt{3}} h=50$

$\therefore \quad h=\dfrac{50 \sqrt{3}}{2}$

$h=25 \sqrt{3} m$

Hence, the required height of tower is $25 \sqrt{3} m$.

Solution

Let the height of the tower be $H$ and $O R=x$

Given that, height of flag staff $=h=F P$ and $\angle P R O=\alpha, \angle F R O=\beta$

Now, in $\triangle P R O, \quad \tan \alpha=\dfrac{P O}{R O}=\dfrac{H}{x}$

$\Rightarrow \quad x=\dfrac{H}{\tan \alpha}$

and in $\triangle F R O$,

$ \begin{aligned} \tan \beta=\dfrac{F O}{R O}=\dfrac{F P+P O}{R O} & \ldots \text{(i)} \end{aligned} $

$ \begin{aligned} \tan \beta = \dfrac{h+H}{x} \\ \Rightarrow \quad x = \dfrac{h+H}{\tan \beta} & \ldots \text{(ii)} \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned} \dfrac{H}{\tan \alpha} = \dfrac{h+H}{\tan \beta} \\ \Rightarrow \quad H \tan \beta = h \tan \alpha+H \tan \alpha \\ \Rightarrow \quad H \tan \beta-H \tan \alpha = h \tan \alpha \\ \Rightarrow \quad H(\tan \beta-\tan \alpha) = h \tan \alpha \Rightarrow H=\dfrac{h \tan \alpha}{\tan \beta-\tan \alpha} \end{aligned} $

Hence the required height of tower is $\dfrac{h \tan \alpha}{\tan \beta-\tan \alpha}$

Hence proved.

Thinking Process

Firstly, we find the value of $(\tan \theta-\sec \theta)$. By using identity $\sec ^{2} \theta-\tan ^{2} \theta=1$ and get the desired result. If $\sec \theta+\tan \theta=a$

$$ \begin{aligned} \Rightarrow \quad \sec \theta-\tan \theta=\dfrac{1}{a} & \ldots \text{(i)} \end{aligned} $$

Solution

Given, $\quad \tan \theta+\sec \theta=l$

[multiply by ( $\sec \theta-\tan \theta$ ) on numerator and denominator LHS]

$ \begin{aligned} & \Rightarrow \quad \dfrac{(\tan \theta+\sec \theta)(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}=l \quad \Rightarrow \quad \dfrac{(\sec ^{2} \theta-\tan ^{2} \theta)}{(\sec \theta-\tan \theta)}=l \\ & \Rightarrow \quad \dfrac{1}{\sec \theta-\tan \theta}=l \quad[\because \sec ^{2} \theta-\tan ^{2} \theta=1] \\ & \Rightarrow \quad \sec \theta-\tan \theta=\dfrac{1}{l} \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} 2 \sec \theta = l+\dfrac{1}{l} \\ \sec \theta = \dfrac{l^{2}+1}{2 l} \end{aligned} $

Hence proved.

Thinking Process

(i) Firstly we will get the value of $\sin \theta \cdot \cos \theta$ with the help of given both equation.

(ii) Now, squaring both sides of the equation $\sin \theta+\cos \theta=p$ and put the value of $\sin \theta \cdot \cos \theta$. Finally simplify it and get the desired result.

Solution

Given that

$ \begin{aligned} \sin \theta+\cos \theta=p & \ldots \text{(i)} \end{aligned} $

and

$ \sec \theta+cosec \theta=q $

$ \begin{matrix} \Rightarrow & \dfrac{1}{\cos \theta}+\dfrac{1}{\sin \theta}=q \\ \Rightarrow & \dfrac{\sin \theta+\cos \theta}{\sin \theta \cdot \cos \theta}=q \\ \Rightarrow & \dfrac{p}{\sin \theta \cdot \cos \theta}=q \\ \Rightarrow & \sin \theta \cdot \cos \theta=\dfrac{p}{q} \\ & \sin \theta+\cos \theta=p \end{matrix} $

$ \because \sec \theta=\dfrac{1}{\cos \theta} \text{ and } cosec \theta=\dfrac{1}{\sin \theta} $

On squaring both sides, we get

$ \begin{aligned} & (\sin \theta+\cos \theta)^{2}=p^{2} \\ & \Rightarrow \quad(\sin ^{2} \theta+\cos ^{2} \theta)+2 \sin \theta \cdot \cos \theta=p^{2} \quad[\because(a+b)^{2}=a^{2}+2 a b+b^{2}] \\ & \Rightarrow \quad 1+2 \sin \theta \cdot \cos \theta=p^{2} \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1] \\ & \begin{matrix} \Rightarrow & 1+2 \cdot \dfrac{p}{q}=p^{2} & \text{ from Eq. (ii)] } \end{matrix} \\ & \Rightarrow \quad q+2 p=p^{2} q \Rightarrow 2 p=p^{2} q-q \\ & \Rightarrow \quad q(p^{2}-1)=2 p \end{aligned} $

Solution

Given that, $a \sin \theta+b \cos \theta=c$

On squaring both sides,

$ \begin{matrix} \Rightarrow & a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+2 a b \sin \theta \cdot \cos \theta=c^{2} & {[\because(x+y)^{2}=x^{2}+2 x y+y^{2}]} \\ \Rightarrow & a^{2}(1-\cos ^{2} \theta)+b^{2}(1-\sin ^{2} \theta)+2 a b \sin \theta \cdot \cos \theta=c^{2} & {[\because \sin ^{2} \theta+\cos ^{2} \theta=1]} \\ \Rightarrow & a^{2}-a^{2} \cos ^{2} \theta+b^{2}-b^{2} \sin ^{2} \theta+2 a b \sin \theta \cdot \cos \theta=c^{2} \\ \Rightarrow & a^{2}+b^{2}-c^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta-2 a b \sin \theta \cdot \cos \theta \\ \Rightarrow & (a^{2}+b^{2}-c^{2})=(a \cos \theta-b \sin \theta)^{2} & {[\because a^{2}+b^{2}-2 a b=(a-b)^{2}]} \\ \Rightarrow & (a \cos \theta-b \sin \theta)^{2}=a^{2}+b^{2}-c^{2} & \\ \Rightarrow & a \cos \theta-b \sin \theta=\sqrt{a^{2}+b^{2}-c^{2}} & \end{matrix} $

Solution

LHS $=\dfrac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}$

$ \begin{aligned} & =\dfrac{1+1 / \cos \theta-\sin \theta / \cos \theta}{1+1 / \cos \theta+\sin \theta / \cos \theta} \quad \because \sec \theta=\dfrac{1}{\cos \theta} \text{ and } \tan \theta=\dfrac{\sin \theta}{\cos \theta} \\ & =\dfrac{\cos \theta+1-\sin \theta}{\cos \theta+1+\sin \theta}=\dfrac{(\cos \theta+1)-\sin \theta}{(\cos \theta+1)+\sin \theta}=\dfrac{2 \cos ^{2} \dfrac{\theta}{2}-2 \sin \dfrac{\theta}{2} \cdot \cos \dfrac{\theta}{2}}{2 \cos ^{2} \dfrac{\theta}{2}+2 \sin \dfrac{\theta}{2} \cdot \cos \dfrac{\theta}{2}} \\ & \because 1+\cos \theta=2 \cos ^{2} \dfrac{\theta}{2} \text{ and } \sin \theta=2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} \\ & =\dfrac{2 \cos ^{2} \dfrac{\theta}{2}-2 \sin \dfrac{\theta}{2} \cdot \cos \dfrac{\theta}{2}}{2 \cos ^{2} \dfrac{\theta}{2}+2 \sin \dfrac{\theta}{2} \cdot \cos \dfrac{\theta}{2}}=\dfrac{2 \cos \dfrac{\theta}{2} \cos \dfrac{\theta}{2}-\sin \dfrac{\theta}{2}}{2 \cos \dfrac{\theta}{2} \cos \dfrac{\theta}{2}+\sin \dfrac{\theta}{2}} \\ & =\dfrac{\cos \dfrac{\theta}{2}-\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}+\sin \dfrac{\theta}{2}} \times \dfrac{\cos \dfrac{\theta}{2}-\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}-\sin \dfrac{\theta}{2}} \\ & =\dfrac{\cos \dfrac{\theta}{2}-\sin \dfrac{\theta}{2}}{\cos ^{2} \dfrac{\theta}{2}-\sin ^{2} \dfrac{\theta}{2}} \quad[\because(a-b)^{2}=a^{2}+b^{2}-2 a b \text{ and }(a-b)(a+b)=(a^{2}-b^{2})] \\ & =\dfrac{\cos ^{2} \dfrac{\theta}{2}+\sin ^{2} \dfrac{\theta}{2}-2 \sin \dfrac{\theta}{2} \cdot \cos \dfrac{\theta}{2}}{\cos \theta} \quad \because \cos ^{2} \dfrac{\theta}{2}-\sin ^{2} \dfrac{\theta}{2}=\cos \theta \\ & =\dfrac{1-\sin \theta}{\cos \theta} \quad \because \sin ^{2} \dfrac{\theta}{2}+\cos ^{2} \dfrac{\theta}{2}=1 \end{aligned} $

$=$ RHS Hence proved.

Solution

Let distance between the two towers $=A B=x m$ and height of the other tower $=P A=h m$

Given that, height of the tower $=Q B=30 m$ and $\angle Q A B=60^{\circ}, \angle P B A=30^{\circ}$

Now, in $\triangle Q A B, \quad \tan 60^{\circ}=\dfrac{Q B}{A B}=\dfrac{30}{x}$

$ \begin{matrix} \Rightarrow & \sqrt{3}=\dfrac{30}{x} \\ \therefore & x=\dfrac{30}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{30 \sqrt{3}}{3}=10 \sqrt{3} m \end{matrix} $

and in $\triangle P B A$,

$ \begin{aligned} \Rightarrow & & \tan 30^{\circ} = \dfrac{P A}{A B}=\dfrac{h}{x} \\ \Rightarrow & & \dfrac{1}{\sqrt{3}} = \dfrac{h}{10 \sqrt{3}} \\ \Rightarrow & & h = 10 m \end{aligned} $

Hence, the required distance and height are $10 \sqrt{3} m$ and $10 m$, respectively.

Solution

Let the distance between two objects is $x m$. and $C D=y m$.

$ \text{ Given that, } \angle BAX =\alpha=\angle ABD \quad \text{[alternate angle]}\\ \hspace{14 mm} \angle CAY = \beta = \angle ACD \quad \text{[alternate angle]} $

and the height of tower, $A D=h m$

Now, in $\triangle A C D$,

$ \begin{aligned} \tan \beta = \dfrac{A D}{C D}=\dfrac{h}{y} \\ \Rightarrow \quad y = \dfrac{h}{\tan \beta} & \ldots \text{(i)} \end{aligned} $

and in $\triangle A B D$,

$ \begin{aligned} & & \tan \alpha=\dfrac{A D}{B D} \Rightarrow=\dfrac{A D}{B C+C D} \\ \Rightarrow & \tan \alpha = \dfrac{h}{x+y} \Rightarrow x+y=\dfrac{h}{\tan \alpha} \\ \Rightarrow & & y=\dfrac{h}{\tan \alpha}-x & \ldots \text{(ii)} \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned}\dfrac{h}{\tan \beta} = \dfrac{h}{\tan \alpha}-x \\ x = \dfrac{h}{\tan \alpha}-\dfrac{h}{\tan \beta} \\ & =h \dfrac{1}{\tan \alpha}-\dfrac{1}{\tan \beta}=h(\cot \alpha-\cot \beta) \quad \because \cot \theta=\dfrac{1}{\tan \theta} \end{aligned} $

which is the required distance between the two objects.

Thinking Process

(i) First, we draw a figure in which the both positions of ladder are shown. In required figure generate two triangles.

(ii) In first position, when ladder makes an angle of elevation is $\alpha$,we use the trigonometric ratios $\sin \theta$ and $\cos \theta$ and get the vertical and horizontal height respectively.

(iii) In second position, when ladder makes an angle of elevation is $\beta$, due to its past is pulled away from the wall, we use again the trigonometric ratios $\sin \theta$ and $\cos \theta$ and get the another vertical and horizontal height respectively. In both case length of ladder is same.

(iv) Now, we find the value of $p$ and $q$ with the help of steps (ii) and (iii) and get the desired result.

Solution

Let $O Q=x$ and $O A=y$

Given that, $\quad B Q=q, S A=P$ and $A B=S Q=$ Length of ladder

Also, $\quad \angle B A O=\alpha$ and $\angle Q S O=\beta$

Now, in $\triangle B A O$,

$ \begin{aligned} & \cos \alpha=\dfrac{O A}{A B} \\ & \Rightarrow \quad \cos \alpha=\dfrac{y}{A B} \\ & \Rightarrow \quad y=A B \cos \alpha=O A \\ & \text{ and } \quad \sin \alpha=\dfrac{O B}{A B} \\ & \Rightarrow \quad O B=B A \sin \alpha \end{aligned} $

Now, in $\triangle Q S O$

$$ \begin{equation*} \cos \beta=\dfrac{O S}{S Q} \tag{ii} \end{equation*} $$

$\Rightarrow \quad O S=S Q \cos \beta=A B \cos \beta$

and

Eq. (v) divided by Eq. (vi), we get

$ \begin{aligned} \dfrac{p}{q} = \dfrac{A B(\cos \beta-\cos \alpha)}{A B(\sin \alpha-\sin \beta)}=\dfrac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta} & \ldots \text{(vi)}\\ \Rightarrow \quad \dfrac{p}{q} = \dfrac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta} \end{aligned} $

$ \begin{aligned} O Q = S Q \sin \beta=A B \sin \beta & \ldots \text{(iv)}\\ S A = O S-A O \\ P = A B \cos \beta-A B \cos \alpha \\ P = A B(\cos \beta-\cos \alpha) & \ldots \text{(v)} \end{aligned} $

$[\because A B=S Q]$

$[\because A B=S Q]$

Hence proved.

Solution

Let the height the vertical tower, $O T=H$ and

$ O P=A B=x m $

Given that,

$A P=10 m$

and

$\angle T P O=60^{\circ}, \angle T A B=45^{\circ}$

Now, in $\triangle T P O$,

$\tan 60^{\circ}=\dfrac{O T}{O P}=\dfrac{H}{x}$

$\Rightarrow$

$\sqrt{3}=\dfrac{H}{x}$

$x=\dfrac{H}{\sqrt{3}}$

and in $\triangle T A B$,

$\tan 45^{\circ}=\dfrac{T B}{A B}=\dfrac{H-10}{x}$

$\Rightarrow$

$1=\dfrac{H-10}{x} \Rightarrow x=H-10$

$\Rightarrow \quad \dfrac{H}{\sqrt{3}}=H-10$

[from Eq. (i)]

$\Rightarrow \quad H-\dfrac{H}{\sqrt{3}}=10 \Rightarrow H 1-\dfrac{1}{\sqrt{3}}=10$

$\Rightarrow \quad H \dfrac{\sqrt{3}-1}{\sqrt{3}}=10$

$ \begin{aligned} & \therefore \quad H=\dfrac{10 \sqrt{3}}{\sqrt{3}-1} \cdot \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\ & =\dfrac{10 \sqrt{3}(\sqrt{3}+1)}{3-1}=\dfrac{10 \sqrt{3}(\sqrt{3}+1)}{2} \\ & =5 \sqrt{3}(\sqrt{3}+1)=5(\sqrt{3}+3) m \end{aligned} $

[by rationalisation]

Hence, the required height of the tower is $5(\sqrt{3}+3) m$.

Solution

Let the height of the other house $=O Q=H$ and

$ O B=M W=x m $

Given that, height of the first house $=W B=h=M O$ and $\angle Q W M=\alpha, \angle O W M=\beta=\angle W O B$

[alternate angle]

Now, in $\triangle W O B, \quad \tan \beta=\dfrac{W B}{O B}=\dfrac{h}{x}$

$$ \begin{equation*} \Rightarrow \quad x=\dfrac{h}{\tan \beta} \tag{i} \end{equation*} $$

And in $\triangle Q W M, \quad \tan \alpha=\dfrac{Q M}{W M}=\dfrac{O Q-M O}{W M}$

$\Rightarrow \quad \tan \alpha=\dfrac{H-h}{x}$

$\Rightarrow \quad x=\dfrac{H-h}{\tan \alpha}$

From Eqs. (i) and (ii),

$ \begin{matrix} \dfrac{h}{\tan \beta} = \dfrac{H-h}{\tan \alpha} \\ \Rightarrow \quad h \tan \alpha = (H-h) \tan \beta \\ \Rightarrow \quad & h \tan \alpha = H \tan \beta-h \tan \beta \\ \Rightarrow \quad H & H \tan \beta = h(\tan \alpha+\tan \beta) \\ \therefore \quad = h \dfrac{\tan \alpha+\tan \beta}{\tan \beta} \\ & =h 1+\tan \alpha \cdot \dfrac{1}{\tan \beta}=h(1+\tan \alpha \cdot \cot \beta) \quad \because \cot \theta=\dfrac{1}{\tan \theta} \end{matrix} $

Hence, the required height of the other house is $h(1+\tan \alpha \cdot \cot \beta)$

Hence proved.

Solution

Let the height of the balloon from above the ground is $H$.

$ A \text{ and } O P=w_2 R=w_1 Q=x $

Given that, height of lower window from above the ground $=w_2 P=2 m=O R$

Height of upper window from above the lower window $=w_1 w_2=4 m=Q R$

$\therefore$

$ \begin{aligned} B Q = O B-(Q R+R O) \\ & =H-(4+2) \\ & =H-6 \end{aligned} $

and

$ \angle B w_1 Q=30^{\circ} $

$\Rightarrow$

$ \angle B w_2 R=60^{\circ} $

Now, in $\Delta B w_2 R$

$$ \begin{align*} \tan 60^{\circ} = \dfrac{B R}{w_2 R}=\dfrac{B Q+Q R}{x} \\ \sqrt{3} = \dfrac{(H-6)+4}{x} \\ x = \dfrac{H-2}{\sqrt{3}} \quad \ldots \text{ (i) } \tag{i} \end{align*} $$

and in $\triangle B w_1 Q$,

$$ \begin{align*} \tan 30^{\circ} = \dfrac{B Q}{W_1 Q} \\ \tan 30^{\circ} = \dfrac{H-6}{x}=\dfrac{1}{\sqrt{3}} \\ \Rightarrow \quad x = \sqrt{3}(H-6) \tag{ii} \end{align*} $$

From Eqs. (i) and (ii),

$ \begin{matrix} \sqrt{3}(H-6) = \dfrac{(H-2)}{\sqrt{3}} \\ \Rightarrow & 3(H-6) = H-2=3 H-18=H-2 \\ \Rightarrow & 2 H = 16 \Rightarrow H=8 \end{matrix} $

So, the required height is $8 m$.

Hence, the required height of the ballon from above the ground is $8 m$.