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 $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$ to get the value of $\tan \theta$.

Solution

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

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

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 $\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$ to get the value of $\cot \theta$.

Solution

(a) Given, $\sin A={\displaystyle \frac{1}{2}}$

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

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

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

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

Solution

(c) Given,

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

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

Solution

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

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

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

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, $\cot \theta ={\displaystyle \frac{1}{\tan \theta }}$.

Solution

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

$\begin{array}{rl}& =\tan 1^{\circ }\cdot \tan 2^{\circ }\cdot \tan 3^{\circ }\dots \tan {44}^{\circ }\cdot \tan {45}^{\circ }\cdot \tan {46}^{\circ }\dots \tan {87}^{\circ }\cdot \tan {88}^{\circ }\cdot \tan {89}^{\circ }\\ & =\tan 1^{\circ }\cdot \tan 2^{\circ }\cdot \tan 3^{\circ }\dots \tan {44}^{\circ }\cdot (1)\cdot \tan ({90}^{\circ }-{44}^{\circ })\dots \tan ({90}^{\circ }-3^{\circ })\cdot \\ & \tan ({90}^{\circ }-2^{\circ })\cdot \tan ({90}^{\circ }-1^{\circ })(\because \tan {45}^{\circ }=1)\\ & =\tan 1^{\circ }\cdot \tan 2^{\circ }\cdot \tan 3^{\circ }\dots \tan {44}^{\circ }\cdot (1)\cdot \cot {44}^{\circ }\dots \dots \cot 3^{\circ }\cdot \cot 2^{\circ }\cdot \cot 1^{\circ }\\ & =[\because \tan ({90}^{\circ }-\theta )=\cot \theta ]\\ & =\tan 1^{\circ }\cdot \tan 2^{\circ }\cdot \tan 3^{\circ }\dots \tan {44}^{\circ }(1)\cdot {\displaystyle \frac{1}{\tan {44}^{\circ }}}\cdots {\displaystyle \frac{1}{\tan {30}^{\circ }}}\cdot {\displaystyle \frac{1}{\tan 2^{\circ }}}\cdot {\displaystyle \frac{1}{\tan 1^{\circ }}}\because \cot \theta ={\displaystyle \frac{1}{\tan \theta }}\\ & =1\end{array}$

Solution

(c) Given,

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

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

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

not in between.

Solution

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

Solution

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

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

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

On squaring both sides, we get

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

Solution

(d) Given,

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

$\begin{array}{rl}& \Rightarrow \alpha ={30}^{\circ }\\ & \text{ and }\cos \beta ={\displaystyle \frac{1}{2}}=\cos {60}^{\circ }\\ & \Rightarrow \beta ={60}^{\circ }\\ & \therefore \alpha +\beta ={30}^{\circ }+{60}^{\circ }={90}^{\circ }\end{array}$

Solution

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

$\begin{array}{rl}& ={\displaystyle \frac{{\sin }^2{22}^{\circ }+{\sin }^2({90}^{\circ }-{22}^{\circ })}{{\cos }^2({90}^{\circ }-{68}^{\circ })+{\cos }^2{68}^{\circ }}}+{\sin }^2{63}^{\circ }+\cos {63}^{\circ }\sin ({90}^{\circ }-{63}^{\circ })\\ & ={\displaystyle \frac{{\sin }^2{22}^{\circ }+{\cos }^2{22}^{\circ }}{{\sin }^2{68}^{\circ }+{\cos }^2{68}^{\circ }}}+{\sin }^2{63}^{\circ }+\cos {63}^{\circ }\cdot \cos {63}^{\circ }\begin{array}{c}[\because \sin ({90}^{\circ }-\theta )=\cos \theta \text{ and }\cos ({90}^{\circ }-\theta )=\sin \theta ]\end{array}\\ & ={\displaystyle \frac{1}{1}}+({\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ })[\because {\sin }^2\theta +{\cos }^2\theta =1]\\ & =1+1=2\end{array}$

Solution

(c) Given,

$4\tan \theta =3$

$\Rightarrow \tan \theta ={\displaystyle \frac{3}{4}}$

$\therefore {\displaystyle \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }}={\displaystyle \frac{4{\displaystyle \frac{\sin \theta }{\cos \theta }}-1}{4{\displaystyle \frac{\sin \theta }{\cos \theta }}+1}}$

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

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

$={\displaystyle \frac{4{\displaystyle \frac{3}{4}}-1}{4{\displaystyle \frac{3}{4}}+1}}={\displaystyle \frac{3-1}{3+1}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}$

[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,

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

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

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

Now,

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

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

$\begin{array}{c}={{\displaystyle \frac{1}{\sqrt{2}}}}^4+{{\displaystyle \frac{1}{\sqrt{2}}}}^4\because \sin {45}^{\circ }=\cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}\\ ={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}\end{array}$

Solution

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

Solution

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

Now, in $\mathrm{△}BAC$,

$\tan \theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow \tan \theta ={\displaystyle \frac{6}{2\sqrt{3}}}={\displaystyle \frac{3}{\sqrt{3}}}\cdot {\displaystyle \frac{\sqrt{3}}{\sqrt{3}}}$

$\Rightarrow \tan \theta ={\displaystyle \frac{3\sqrt{3}}{3}}=\sqrt{3}=\tan {60}^{\circ }$

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

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

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

Solution

True

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

Solution

False

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

which may be either positive or negative.

Solution

False

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

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

[positive]

Solution

True

$\begin{array}{rl}& \sqrt{(1-{\cos }^2\theta ){\sec }^2\theta }=\sqrt{{\sin }^2\theta \cdot {\sec }^2\theta }[\because {\sin }^2\theta +{\cos }^2\theta =1]\\ & =\sqrt{{\sin }^2\theta \cdot {\displaystyle \frac{1}{{\cos }^2\theta }}}=\sqrt{{\tan }^2\theta }=\tan \theta \because \sec \theta ={\displaystyle \frac{1}{\cos \theta }},\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}\end{array}$

Solution

True

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

Solution

False

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

Solution

False

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

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

$\begin{array}{ccc}\text{ In }\mathrm{△}ACB,& & \tan \theta ={\displaystyle \frac{AB}{BC}}={\displaystyle \frac{2\sqrt{3}}{2}}\\ \Rightarrow & \tan \theta =\sqrt{3}=\tan {60}^{\circ }\\ \therefore & & \theta ={60}^{\circ }\end{array}$

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

$\begin{array}{rl}& \text{ In }\mathrm{△}APB,\tan \theta ={\displaystyle \frac{AB}{PB}}={\displaystyle \frac{AB}{PC+CB}}\\ & \Rightarrow \tan \theta ={\displaystyle \frac{2\sqrt{3}}{4+2}}={\displaystyle \frac{2\sqrt{3}}{6}}\\ & \Rightarrow \tan \theta ={\displaystyle \frac{\sqrt{3}}{3}}\cdot {\displaystyle \frac{\sqrt{3}}{\sqrt{3}}}={\displaystyle \frac{3}{3\sqrt{3}}}\\ & \Rightarrow \tan \theta ={\displaystyle \frac{1}{\sqrt{3}}}=\tan {30}^{\circ }\\ & \therefore \theta =30\end{array}$

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 $P,3m$ 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.

$\begin{array}{cc}\text{ In }\mathrm{△}MPC,& \tan {\theta }_1={\displaystyle \frac{CM}{PM}}={\displaystyle \frac{h}{PM}}\\ \Rightarrow & {\displaystyle \frac{\tan {\theta }_1}{h}}={\displaystyle \frac{1}{PM}}& \dots \text{(ii)}\\ \text{ In }\mathrm{△}CPM,& \tan {\theta }_2={\displaystyle \frac{CM}{PM}}={\displaystyle \frac{OC}{F}}\\ \Rightarrow & {\displaystyle \frac{\tan {\theta }_2}{h+3}}={\displaystyle \frac{1}{PM}}\end{array}$

Hence,

Alternate Method

${\displaystyle \frac{\tan {\theta }_1}{h}}={\displaystyle \frac{\tan {\theta }_2}{h+3}}$

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 $3m$ 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 $a\ne 1$, then $AM>GM$

$\Rightarrow {\displaystyle \frac{a+{\displaystyle \frac{1}{a}}}{2}}>\sqrt{a\cdot {\displaystyle \frac{1}{a}}}\Rightarrow a+{\displaystyle \frac{1}{a}}>2$

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

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

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

which is not possible.

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

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

Solution

False

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

Using,

$AM>GM$

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

$\Rightarrow {\displaystyle \frac{a^2+b^2}{2}}>\sqrt{a^2\cdot b^2}$

$\Rightarrow a^2+b^2>2ab$

$\Rightarrow {\displaystyle \frac{a^2+b^2}{2ab}}>1\because \cos \theta ={\displaystyle \frac{a^2+b^2}{2ab}}$

$\Rightarrow \cos \theta >1[\because -1\le \cos \theta \le 1]$

which is not possible.

Hence, $\cos \theta \ne {\displaystyle \frac{a^2+b^2}{2ab}}$

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 $BC=xm$ In $\mathrm{△}ABC$,

$\begin{array}{rl}& \tan {30}^{\circ }={\displaystyle \frac{AC}{BC}}={\displaystyle \frac{h}{x}}\\ & \Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{x}}\end{array}$

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

$\begin{array}{ccc}\text{ In }\mathrm{△}PQR,& & \tan \theta ={\displaystyle \frac{PR}{QR}}={\displaystyle \frac{2h}{x}}\\ \Rightarrow & & \tan \theta ={\displaystyle \frac{2}{x}}\times {\displaystyle \frac{x}{\sqrt{3}}}\\ \Rightarrow & & \tan \theta ={\displaystyle \frac{2}{\sqrt{3}}}=1.15\\ & \therefore & \theta ={\tan }^{-1}(1.15)<{60}^{\circ }\end{array}$

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

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 $\mathrm{△}ABC$,

$\begin{array}{r}\tan {\theta }_1={\displaystyle \frac{AC}{BC}}={\displaystyle \frac{h}{x}}\\ \Rightarrow {\theta }_1={\tan }^{-1}{\displaystyle \frac{h}{x}}& \dots \text{(i)}\end{array}$

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

$\begin{array}{cc}& =x+x\times {\displaystyle \frac{10}{100}}={\displaystyle \frac{11x}{10}}\\ \ln \mathrm{△}PQR,\tan {\theta }_2={\displaystyle \frac{PR}{QR}}={\displaystyle \frac{{\displaystyle \frac{11h}{10}}}{{\displaystyle \frac{11x}{10}}}}\\ \Rightarrow & \tan {\theta }_2={\displaystyle \frac{h}{x}}\\ \Rightarrow & {\theta }_2={\tan }^{-1}{\displaystyle \frac{h}{x}}& \dots \text{(ii)}\end{array}$

From Eqs. (i) and (ii),

${\theta }_1={\theta }_2$

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

Solution

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

$\begin{array}{cc}={\displaystyle \frac{{\sin }^2\theta +1+{\cos }^2\theta +2\cos \theta }{\sin \theta (1+\cos \theta )}}& [\because (a+b{)}^2=a^2+b^2+2ab]\\ ={\displaystyle \frac{1+1+2\cos \theta }{\sin \theta (1+\cos \theta )}}& \\ ={\displaystyle \frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}}={\displaystyle \frac{2}{\sin \theta }}& \\ =2cosec\theta =\text{ RHS }& \end{array}$

Solution

$LHS={\displaystyle \frac{\tan A}{1+\sec A}}-{\displaystyle \frac{\tan A}{1-\sec A}}={\displaystyle \frac{\tan A(1-\sec A-1-\sec A)}{(1+\sec A)(1-\sec A)}}$

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

Thinking Process

We know that, $\tan \theta ={\displaystyle \frac{\text{ Perpendicular }}{\text{ Base }}}$. 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,

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

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

By Pythagoras theorem,

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

Hence proved.

Solution

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

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

Solution

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

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

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

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

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

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

Hence proved.

Solution

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

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

Solution

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

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