Solution

$ \begin{aligned} LHS & =\frac{\tan A+\sec A-1}{\tan A-\sec A+1} \\ & =\frac{\tan A+\sec A-(\sec ^{2} A-\tan ^{2} A)}{(\tan A-\sec A+1)} \\ \\ & [\because \sec ^{2} A-\tan ^{2} A=1] \\ \\ & =\frac{(\tan A+\sec A)-(\sec A+\tan A)(\sec A-\tan A)}{(1-\sec A+\tan A)} \\ & =\frac{(\sec A+\tan A)(1-\sec A+\tan A)}{1-\sec A+\tan A} \\ & =\sec A+\tan A=\frac{1}{\cos A}+\frac{\sin A}{\cos A} \\ & =\frac{1+\sin A}{\cos A}=RHS \quad \text { Hence proved. } \end{aligned} $

Solution

Given that, $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y$

$ \begin{aligned} \text { Now, } & \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \\ & =\frac{(1-\cos \alpha+\sin \alpha)}{(1+\sin \alpha)} \cdot \frac{(1+\cos \alpha+\sin \alpha)}{(1+\cos \alpha+\sin \alpha)} \\ & =\frac{{(1+\sin \alpha)-\cos \alpha}}{(1+\sin \alpha)} \cdot \frac{{(1+\sin \alpha)+\cos \alpha}}{(1+\cos \alpha+\sin \alpha)} \\ & =\frac{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{(1+\sin ^{2} \alpha+2 \sin \alpha)-\cos ^{2} \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \end{aligned} $

$ \begin{aligned} & =\frac{1+\sin ^{2} \alpha+2 \sin \alpha-1+\sin ^{2} \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin ^{2} \alpha+2 \sin \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin \alpha}{1+\sin \alpha+\cos \alpha}=y \end{aligned} $

Hence proved.

Solution

Given that,

$ \begin{aligned} & \text { Given that, } \quad \begin{aligned} m \sin \theta & =n \sin (\theta+2 \alpha) \\ \therefore & \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{m}{n} \end{aligned} \end{aligned} $

Using componendo and dividendo, we get

$ \begin{aligned} \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n} \\ \\ \Rightarrow \quad \frac{2 \sin \frac{\theta+2 \alpha+\theta}{2} \cdot \cos \frac{\theta+2 \alpha-\theta}{2}}{2 \cos \frac{\theta+2 \alpha+\theta}{2} \cdot \sin \frac{\theta+2 \alpha-\theta}{2}}=\frac{m+n}{m-n} \\ \\ \Rightarrow \quad[\because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}.\\ \text { and } .\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \cdot \frac{x-y}{2}] \\ \\ \Rightarrow \quad \frac{\sin (\theta+\alpha) \cdot \cos \alpha}{\cos (\theta+\alpha) \cdot \sin \alpha}=\frac{m+n}{m-n} \\ \\ \tan (\theta+\alpha) \cdot \cot \alpha =\frac{m+n}{m-n} \quad \text { Hence proved. } \end{aligned} $

Solution

Given that,

$ \cos (\alpha+\beta)=\frac{4}{5} \text { and } \sin (\alpha-\beta)=\frac{5}{13} $

$\Rightarrow \sin (\alpha+\beta)=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}= \pm \frac{3}{5}$

$\therefore \sin (\alpha+\beta)=\frac{3}{5}$

$\text { and } \cos (\alpha-\beta)=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}= \pm \frac{12}{13}$

$\therefore \cos (\alpha-\beta)=\frac{12}{13}$

$\text { Now, } \quad \tan (\alpha+\beta)=\frac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)} $

$[\text { since, } \alpha\text { lies between } 0 \text { and } \frac{\pi}{4}]$

$ \quad =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

$ \text { and } \quad \tan (\alpha-\beta) =\frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12} $

$ \therefore \quad \tan 2 \alpha =\tan (\alpha+\beta+\alpha-\beta)$

$ \qquad =\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \quad [ \because \tan (x \pm y)=\frac{\tan x \pm \tan y}{1 \mp \tan x \cdot \tan y} ]$

$ \qquad =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}=\frac{\frac{9+5}{12}}{\frac{16-5}{16}}=\frac{14 \times 16}{12 \times 11}=\frac{56}{33} $

Solution

Given that, $\quad \tan x=\frac{b}{a}$

$ \therefore \quad \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} =\frac{\sqrt{(a+b)^{2}}+\sqrt{(a-b)^{2}}}{\sqrt{(a-b)(a+b)}} $

$=\frac{(a+b)+(a-b)}{\sqrt{a^{2}-b^{2}}}=\frac{2 a}{\sqrt{a^{2}-b^{2}}}=\frac{2 a}{a \sqrt{1-\frac{b}{a}}} \quad [\because \frac{b}{a}=\tan ]$

$=\frac{2}{\sqrt{1-\tan ^{2} x}}=\frac{2 \cos x}{\sqrt{\cos ^{2} x-\sin ^{2} x}} \quad [\because \cos 2 x=\cos ^{2} x-\sin ^{2} x] x $

$ \quad =\frac{2 \cos x}{\sqrt{\cos 2 x}} $

Solution

LHS $=\cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}$

$ \begin{aligned} & =\frac{1}{2} 2 \cos \theta \cdot \cos \frac{\theta}{2}-2 \cos 3 \theta \cdot \cos \frac{9 \theta}{2} \\ \\ & =\frac{1}{2} \cos \theta+\frac{\theta}{2}+\cos \theta-\frac{\theta}{2}-\cos 3 \theta+\frac{9 \theta}{2}-\cos 3 \theta-\frac{9 \theta}{2} \\ \\ & =\frac{1}{2}(\cos \frac{3 \theta}{2}+\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}. \\ \\ & =\frac{1}{2} \cos \frac{\theta}{2}-\cos \frac{15 \theta}{2} \end{aligned} $

$ =-\frac{1}{2} 2 \sin \frac{\theta+15 \theta}{2} \cdot \sin \frac{\theta-15 \theta}{2} $

$[\because \cos x-\cos y=-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}] $

$ =+(\sin 8 \theta \cdot \sin 7 \theta)=RHS $

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

Hence Proved

Solution

Given that,

and

$a \cos \theta+b \sin \theta=m \qquad ..(i)$

$a \sin \theta-b \cos \theta=n \qquad ..(ii)$

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

$ m^{2}+n^{2}=(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2} $

$= a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cdot \cos \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta $

$\Rightarrow \quad m^{2}+n^{2}=a^{2}(\cos ^{2} \theta+\sin ^{2} \theta)+b^{2}(\sin ^{2} \theta+\cos ^{2} \theta) -2 a b \sin \theta \cdot \cos \theta $

$\Rightarrow \quad m^{2}+n^{2}=a^{2}+b^{2} \quad \text { Hence proved. } $

Solution

Let

$\text { Let }\theta =45^{\circ} $

$\text { We know that, } \quad \tan \frac{\theta}{2} =\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \Rightarrow \tan \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta} $

$\therefore \quad \tan 22^{\circ} 30^{\prime} =\frac{\sin 45^{\circ}}{1+\cos 45^{\circ}} \quad[\because \theta=45^{\circ}]$

$ =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1} $

Solution

$ LHS =\sin 4 A $

$ =2 \sin 2 A \cdot \cos 2 A $

$ =2(2 \sin A \cdot \cos A)(\cos ^{2} A-\sin ^{2} A) $

$ =4 \sin A \cdot \cos ^{3} A-4 \cos A \sin ^{3} A $

$[\because \cos 2 A=\cos ^{2} A-\sin ^{2} A \quad \text { and } \sin 2 A=2 \sin A \cdot \cos A]$

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

Solution

Now,

Also, Given that,

$\quad \tan \theta+\sin \theta =m \qquad ..(i) $

$\qquad \text {and} \quad \tan \theta-\sin \theta =n \qquad ..(ii)$

$\qquad \text {Now} \quad m+n =\tan \theta+\sin \theta+\tan \theta-\sin \theta $

$\qquad m+n =2 \tan \theta \qquad …(iii) $

$\text {also} \qquad m-n =\tan \theta+\sin \theta-\tan \theta+\sin \theta $

$ m-n =2 \sin \theta \quad …(iv) $

From Eqs. (iii) and (iv),

$ \begin{aligned} (m+n)(m-n) & =4 \sin \theta \cdot \tan \theta \\ m^{2}-n^{2} & =4 \sin \theta \cdot \tan \theta \qquad \text{Hence Proved} \end{aligned} $

Solution

Given that $ \quad \tan (A+B)=p \qquad ..(i)$

and $\quad \tan (A-B)=q \qquad ..(ii)$

$\therefore \quad \tan 2 A=\tan (A+B+A-B)$

$=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)} \quad[ \because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}]$

$=\frac{p+q}{1-p q} $ $\qquad$ [from Eqs. (i) and (ii)]

Solution

Given that, $\quad \cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$

$\Rightarrow \quad(\cos \alpha+\cos \beta)^{2}-(\sin \alpha+\sin \beta)^{2}=0$

$\Rightarrow \quad \cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta-\sin ^{2} \alpha-\sin ^{2} \beta-2 \sin \alpha \sin \beta=0$

$\Rightarrow \quad \cos ^{2} \alpha-\sin ^{2} \alpha+\cos ^{2} \beta-\sin ^{2} \beta=2(\sin \alpha \sin \beta-\cos \alpha \cos \beta)$

$\Rightarrow \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)$

Hence proved.

Solution

Given that, $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$

Using componendo and dividendo,

$ \Rightarrow =\frac{\sin (x+y)+[\sin (x-y)]}{\sin (x+y)-\sin (x-y)}=\frac{a+b+a-b}{a+b-a+b} $

$\Rightarrow =\frac{2 \sin \frac{x+y+x-y}{2} \cdot \cos \frac{x+y-x+y}{2}}{2 \cos \frac{x+y+x-y}{2} \cdot \sin \frac{x+y-x+y}{2}}=\frac{2 a}{2 b}$

$[ \because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} \text { and } \sin x-\sin y=2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}] $

$\Rightarrow =\frac{\sin x \cdot \cos y}{\cos x \cdot \sin y}=\frac{a}{b} $

$ \frac{\tan x}{\tan y}=\frac{a}{b} $

Solution

Given that, $\quad \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$

$ \begin{matrix} \Rightarrow \qquad \tan \theta=\frac{\cos \alpha(\tan \alpha-1)}{\cos \alpha(\tan \alpha+1)} & \\ \\ \Rightarrow \quad \tan \theta=\frac{\tan \alpha-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{4} \cdot \tan \alpha} & [\because \tan \frac{\pi}{4}=1] \end{matrix} $

Trigonometric Functions

$ \Rightarrow \quad \tan \theta=\tan (\alpha-\frac{\pi}{4}) $

$\Rightarrow \quad \theta=\alpha-\frac{\pi}{4} \Rightarrow \alpha=\theta+\frac{\pi}{4} $

$\therefore \quad \sin \alpha+\cos \alpha=\sin (\theta+\frac{\pi}{4})+\cos (\theta+\frac{\pi}{4})$

$=\sin \theta \cdot \cos \frac{\pi}{4}+\cos \theta \cdot \sin \frac{\pi}{4}+\cos \theta \cdot \cos \frac{\pi}{4}-\sin \theta \cdot \sin \frac{\pi}{4} $

$=\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \quad [\because \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}]$

$ =\frac{2}{\sqrt{2}} \cdot \cos \theta=\sqrt{2} \cos \theta $

Solution

Given that, $\sin \theta+\cos \theta=1$

On squaring both sides, we get

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

$\Rightarrow \qquad 1+2 \sin \theta \cdot \cos =1 \quad [\because \sin 2 x=2 \sin x \cos x] $

$\Rightarrow \qquad \sin 2 \theta =0 2 \theta=n \pi+(-1)^{n} \cdot 0 $

$ \therefore \qquad \theta =\frac{n \pi}{2}$

Alternate Method

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

$ \Rightarrow \frac{1}{\sqrt{2}} \cdot \sin \theta+\frac{1}{\sqrt{2}} \cdot \cos \theta=\frac{1}{\sqrt{2}} $

$\Rightarrow \sin \theta \cdot \cos \frac{\pi}{4}+\cos \theta \cdot \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \qquad \because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4} $

$ \Rightarrow \quad \sin \theta+\frac{\pi}{4}=\sin \frac{\pi}{4} \qquad[\because \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y] $

$ \Rightarrow \quad \theta+\frac{\pi}{4}=n \pi+(-1)^{n} \frac{\pi}{4} $

$ \therefore \quad \quad \theta=n \pi+(-1)^{n} \frac{4}{4}-\frac{\pi}{4} $

Solution

The given equations are

$ \tan \theta=-1 \qquad ..(i) $

$ \text { and } \quad \cos \theta=\frac{1}{\sqrt{2}} \qquad ..(ii) $

$ \text { From Eq. (i), } \quad \tan \theta=-\tan \frac{\pi}{4} $

$ \Rightarrow \quad \tan \theta=\tan( 2 \pi-\frac{\pi}{4}) \quad \Rightarrow \tan \theta=\tan \frac{7 \pi}{4} $

$ \therefore \quad \theta=\frac{7 \pi}{4}$

From Eq. (ii),

$ \Rightarrow \quad \cos \theta=\cos (2 \pi-\frac{\pi}{4}) \Rightarrow \cos \theta=\cos \frac{7 \pi}{4} $

$\therefore \quad \theta=\frac{7 \pi}{4}$

Hence, the most general value of $\theta$ i.e., $\theta=2 n \pi+\frac{7 \pi}{4}$.

Solution

Given that,

$ \cot \theta+\tan \theta=2 cosec \theta $

$ \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{2}{\sin \theta} $

$\Rightarrow \frac{\cos ^{2}+\sin ^{2} \theta}{\sin \theta \cdot \cos \theta} =\frac{2}{\sin \theta} $

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

$\Rightarrow \cos \theta =\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} $

$\therefore \qquad \theta =2 n \pi \pm \frac{\pi}{3} $

Solution

Given that,

$\text {Given that ,} \qquad 2 \sin^{2} \theta = 3 \cos \theta $

$ \Rightarrow \qquad 2-2 \cos ^{2} \theta=3 \cos \theta $

$\Rightarrow \qquad 2 \cos ^{2} \theta+3 \cos \theta-=0 $

$\Rightarrow \qquad 2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 $

$ \Rightarrow \qquad 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0 $

$\Rightarrow \qquad (\cos \theta+2)(2 \cos \theta-1)=0 $

$\Rightarrow \qquad \cos \theta=-2 not possible \qquad [\because -1 \leq \cos \theta \leq 1 ] $

$\Rightarrow \qquad 2 \cos \theta=1 $

$\Rightarrow \qquad \cos \theta=\frac{1}{2} $

$\Rightarrow \qquad \cos \theta=\cos \frac{\pi}{3} $

$\therefore \theta=\frac{\pi}{3} $

$Also, \qquad \cos \theta=\cos(2 \pi -\frac{\pi}{3}) $

$\Rightarrow \qquad \cos \theta =\cos \frac{5 \pi}{6} $

$\therefore \qquad \theta=\frac{5 \pi}{6}$

So, the values of $\theta$ are $\frac{\pi}{3}$ and $\frac{5 \pi}{6}$.

Solution

Given that, $\quad \sec x \cos 5 x+1=0$

$\frac{\cos 5 x}{\cos x}+1=0 \Rightarrow \cos 5 x+\cos x=0 $

$ \Rightarrow \quad 2 \cos( \frac{5 x+x}{2}) \cdot \cos (\frac{5 x-x}{2})=0$

$\quad [\because \cos x+\cos y=2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}] $

$ \Rightarrow \quad 2 \cos 3 x \cdot \cos 2 x=0 $

$ \Rightarrow \quad \cos 3 x=0 \text { or } \cos 2 x=0 $

$ \Rightarrow \quad \cos 3 x=\cos \frac{\pi}{2} \text { or } \cos 2 x=\cos \frac{\pi}{2} $

$ \therefore \quad 3 x=\frac{\pi}{2} \Rightarrow 2 x=\frac{\pi}{2} $

$ x=\frac{\pi}{6} \Rightarrow x=\frac{\pi}{4} $

Hence, the solutions are $\frac{\pi}{2}, \frac{\pi}{4}$ and $\frac{\pi}{6}$.

Solution

Given that, $\sin (\theta+\alpha)=a \qquad ..(i)$

and

$\sin (\theta+\beta)=b \qquad ..(ii)$

$\therefore \quad \cos (\theta+\alpha)=\sqrt{1-a^{2}}$ and $\cos (\theta+\beta)=\sqrt{1-b^{2}}$

$\therefore \quad \cos (\alpha-\beta)=\cos {\theta+\alpha-(\theta+\beta)}$

$ =\cos (\theta+\beta) \cos (\theta+\alpha)+\sin (\theta+\alpha) \sin (\theta+\beta) $

$ \begin{aligned} & =\sqrt{1-a^{2}} \sqrt{1-b^{2}}+a \cdot b=a b+\sqrt{(1-a^{2})(1-b^{2})} \\ & =a b+\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}} \end{aligned} $

and

$ \cos (\alpha-\beta) =a b+\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}} $

$ =\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta) $

$ =2 \cos ^{2}(\alpha-\beta)-1-4 a b \cos (\alpha-\beta) $

$ =2 \cos (\alpha-\beta)(\cos \alpha-\beta-2 a b)-1 $

$ =2(a b+\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}})(a b+\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}}-2 a b)-1 $

$ =2[(\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}+a b})(\sqrt{1-a^{2}-b^{2}+a^{2} b^{2}}-a b)]-1 $

$ =2[1-a^{2}-b^{2}+a^{2} b^{2}-a^{2} b^{2}]-1 $

$ =2-2 a^{2}-2 b^{2}-1 $

$ =1-2 a^{2}-2 b^{2} \qquad Hence \quad Proved $

Solution

Given that,

$ \begin{aligned} & \cos (\theta+\varphi)=m \cos (\theta-\varphi) \\ &\Rightarrow \qquad \frac{\cos (\theta+\varphi)}{\cos (\theta-\varphi)}=\frac{m}{1} \end{aligned} $

Using componendo and dividendo rule,

$ \frac{\cos (\theta-\varphi)-\cos (\theta+\varphi)}{\cos (\theta-\varphi)+\cos (\theta+\varphi)} =\frac{1-m}{1+m} $

$\Rightarrow \frac{-2 \sin \frac{\theta-\varphi+\theta+\varphi}{2} \cdot \sin \frac{\theta-\varphi-\theta-\varphi}{2}}{2 \cos \frac{\theta-\varphi+\theta+\varphi}{2} \cdot \cos \frac{\theta-\varphi-\theta-\varphi}{2}} =\frac{1-m}{1+m} $

$\Rightarrow \qquad \frac{\sin \theta \cdot \sin \varphi}{\cos \theta \cdot \cos \varphi} =\frac{1-m}{1+m} \qquad [\because \sin(-\theta)= -\sin \theta and \cos (-\theta)=\cos \theta] $

$\Rightarrow \qquad \tan \theta \cdot \tan \varphi =\frac{1-m}{1+m} $

$\Rightarrow \tan \theta = \frac{1-m}{1+m} \cot \varphi $

Solution

Given expression,

$ \begin{aligned} 3 [\sin ^{4} & (\frac{3 \pi}{2})-\alpha+\sin ^{4}(3 \pi+\alpha)]-2 [\sin ^{6} \frac{\pi}{2}+\alpha+\sin ^{6}(5 \pi-\alpha)] \\ & =3[\cos ^{4} \alpha+\sin ^{4}(\pi+\alpha)]-2[\cos ^{6} \alpha+\sin ^{6}(\pi-\alpha)] \\ & =3[\cos ^{4} \alpha+\sin ^{4} \alpha]-2[\cos ^{6} \alpha+\sin ^{6} \alpha]=3-2=1 \end{aligned} $

Solution

Given that, $a \cos 2 \theta+b \sin 2 \theta=c$

$ \Rightarrow \quad a \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}+b \frac{2 \tan \theta}{1+\tan ^{2} \theta}=c \quad $

$[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta} \text { and } \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}] $

$ \Rightarrow \quad a(1-\tan ^{2} \theta)+2 b \tan \theta=c(1+\tan ^{2} \theta) $

$ \Rightarrow \quad a-a-\tan ^{2} \theta+2 b \tan \theta=c+c \tan ^{2} \theta $

$ \Rightarrow \quad(a+c) \tan ^{2} \theta-2 b \tan \theta+c-a=0 $

Since, this equation has $\tan \alpha$ and $\tan \beta$ as its roots.

$ \because \quad \tan \alpha+\tan \beta=\frac{-(-2 b)}{a+c}=\frac{2 b}{a+c} $

Solution

Given that, and $x=\sec \varphi-\tan \varphi \qquad ..(i)$

Now,

$y=cosec \varphi+\cot \varphi \qquad ..(ii)$

$\Rightarrow \quad x y=\sec \varphi \cdot cosec \varphi-cosec \varphi \cdot \tan \varphi+\sec \varphi \cdot \cot \varphi-\tan \varphi \cdot \cot \varphi$

$\Rightarrow \quad x y=\sec \varphi \cdot cosec \varphi-\frac{1}{\cos \varphi}+\frac{1}{\sin \varphi}-1$

$\Rightarrow \quad 1+x y=\sec \varphi cosec \varphi-\sec \varphi+cosec \varphi \qquad ..(iii)$

From Eqs. (i) and (ii), we get

$ x-y=\sec \varphi-\tan \varphi-cosec \varphi-\cot \varphi $

$\Rightarrow \qquad x-y=\sec \varphi-cosec \varphi-\frac{\sin \varphi}{\cos \varphi}-\frac{\cos \varphi}{\sin \varphi} $

$\Rightarrow \qquad x-y=\sec \varphi-cosec \varphi-(\frac{\sin ^{2} \varphi+\cos ^{2} \varphi}{\sin \varphi \cdot \cos \varphi}) $

$ \Rightarrow \qquad x-y=\sec \varphi-cosec \varphi-\frac{1}{\sin \varphi \cdot \cos \varphi} $

$ \begin{aligned} & \Rightarrow \quad x-y=\sec \varphi-cosec \varphi-cosec \varphi \cdot \sec \varphi \\ & \Rightarrow \quad x-y=-(\sec \varphi \cdot cosec \varphi-\sec \varphi+cosec \varphi) \\ & \Rightarrow \quad x-y=-(x y+1) \\ & \Rightarrow \quad x y+x-y+1=0 \quad \text{[from Eq. (iii)] Hence proved.} \end{aligned} $

Solution

Given that,

$ \Rightarrow \qquad \sin \theta=\sqrt{\frac{289-64}{289}}$

$\Rightarrow \qquad \sin \theta= \pm \frac{15}{17} $

$\Rightarrow \quad \sin \theta=\frac{15}{17} \qquad [\text {Since,} \theta \text {lies in first quadrant }] $

$\text { Now, } \cos (30^{\circ}+\theta)+ \cos (45^{\circ}-\theta)+\cos (120^{\circ}-\theta) $

$= \qquad \cos (30^{\circ}+\theta)+\cos (45^{\circ}-\theta)+\cos (90^{\circ}+30^{\circ}-\theta) $

$= \qquad \cos (30^{\circ}+\theta)+\cos (45^{\circ}-\theta)-\sin (30^{\circ}-\theta) $

$= \qquad \cos 30^{\circ} \cos \theta-\sin 30^{\circ} \sin \theta+\cos 45^{\circ} \cos \theta+\sin 45^{\circ} \sin \theta $

$= \qquad \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{2} \cos \theta \frac{\sqrt{3}}{2} \sin \theta $

$= \qquad \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2} \cos \theta+\frac{1}{\sqrt{2}}-\frac{1}{2}+\frac{\sqrt{3}}{2} \sin \theta $

$= \qquad \frac{\sqrt{6}+2-\sqrt{2}}{2 \sqrt{2}} \cos \theta+\frac{2-\sqrt{2}+\sqrt{6}}{2 \sqrt{2}} \sin \theta $

$= \qquad \frac{\sqrt{6}+2-\sqrt{2}}{2 \sqrt{2}} \frac{8}{17}+\frac{2-\sqrt{2}+\sqrt{6}}{2 \sqrt{2}} \frac{15}{17} $

$ \begin{aligned} & =\frac{1}{17(2 \sqrt{2})}(8 \sqrt{6}+16-8 \sqrt{2}+30-15 \sqrt{2}+15 \sqrt{6}) \\ & =\frac{1}{17(2 \sqrt{2})}(23 \sqrt{6}-23 \sqrt{2}+46) \\ & =\frac{23 \sqrt{6}}{17(2 \sqrt{2})}-\frac{23 \sqrt{2}}{17(2 \sqrt{2})}+\frac{46}{17(2 \sqrt{2})} \\ & =\frac{23 \sqrt{3}}{17(2)}-\frac{23}{17(2)}+\frac{23}{17 \sqrt{2}} \\ & =\frac{23}{17} \frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}} \end{aligned} $

Solution

Given expression, $\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8}$

$ \begin{aligned} & =\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} \pi-\frac{3 \pi}{8}+\cos ^{4} \pi-\frac{\pi}{8} \\ & =\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} (\frac{3 \pi}{8})+\cos ^{4} (\frac{\pi}{8}) \\ & =2 [\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}]=2 [\cos ^{4} \frac{\pi}{8}+\cos ^{4} (\frac{\pi}{2}-\frac{\pi}{8})] \\ & =2 [\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{\pi}{8}] \\ & =2[ \cos ^{2} \frac{\pi}{8}+\sin ^{2} \frac{\pi^{2}}{8}-2 \cos ^{2} \frac{\pi}{8} \cdot \sin ^{2} \frac{\pi}{8}] \\ & =21-2 \cos ^{2} \frac{\pi}{8} \cdot \sin ^{2} \frac{\pi}{8}=2-2 \sin \frac{\pi}{8} \cdot \cos \frac{\pi^{2}}{8} \\ & =2-\sin (\frac{2 \pi}{8})^{2}=2-(\frac{1}{\sqrt{2}}) \\ & =2-\frac{1}{2}=\frac{3}{2} \end{aligned} $

Solution

Given equation,

$ \begin{matrix} \Rightarrow \qquad 5 \cos ^{2} \theta+7 \sin ^{2} \theta-6 & =0 & & \\ \Rightarrow \qquad 5 \cos ^{2}+7(1-\cos ^{2} \theta)-6 & =0 & & \\ \Rightarrow \qquad 5 \cos ^{2} \theta+7-7 \cos ^{2} \theta-6 & =0 & & \\ \Rightarrow \qquad 5 \cos ^{2} \theta+7-7 \cos ^{2} \theta-6 & =0 \Rightarrow & \Rightarrow-2 \cos ^{2} \theta+1=0 \\ \Rightarrow \qquad 2 \cos ^{2} \theta-1 =0 & & [\because \cos ^{2} \theta=\cos ^{2} \alpha] \\ \Rightarrow \qquad \cos ^{2} \theta =\frac{1}{2} & & \therefore \theta=n \pi \pm \alpha \\ \Rightarrow \qquad \cos ^{2} \theta & =\cos ^{2} \frac{\pi}{4} \\ \theta & =n \pi \pm \frac{\pi}{4} & & \end{matrix} $

Solution

Given equation, $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$

$ \begin{aligned} & \Rightarrow \quad 2 \sin (\frac{x+3 x}{2} )\cdot \cos (\frac{3 x-x}{2})-3 \sin 2 x \\ & =2 \cos (\frac{3 x+x}{2}) \cdot \cos (\frac{3 x-x}{2})-3 \cos 2 x \\ & \Rightarrow \quad 2 \sin 2 x \cos x-3 \sin 2 x=2 \cos 2 x \cdot \cos x-3 \cos 2 x \\ & \Rightarrow \quad \sin 2 x(2 \cos x-3)=\cos 2 x(2 \cos x-3) \\ & \Rightarrow \quad \frac{\sin 2 x}{\cos 2 x}=1 \\ & \begin{matrix} \Rightarrow & \tan 2 x=1 \end{matrix} \\ & \Rightarrow \quad \tan 2 x=\tan \frac{\pi}{4} \\ & \Rightarrow \quad 2 x=n \pi+\frac{\pi}{4} \\ & \therefore \quad x=\frac{n \pi}{2}+\frac{4}{8} \end{aligned} $

Solution

Given equation is,

$ (\sqrt{3}-1) \cos \theta+(\sqrt{3}+1) \sin \theta=2 \qquad ..(i)$

$ \begin{aligned} & \text { Put } \quad \sqrt{3}-1=r \sin \alpha \quad \text { and } \quad \sqrt{3}+1=r \cos \alpha \\ & \therefore \quad r^{2}=(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2} \\ & \Rightarrow \quad=3+1-2 \sqrt{3}+3+1+2 \sqrt{3} \\ & \Rightarrow \quad r^{2}=8 \\ & \therefore \quad r=2 \sqrt{2} \\ & \text { now, } \quad \tan \alpha=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\tan \frac{\pi}{3}-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{3} \cdot \frac{\pi}{4}} \\ & \Rightarrow \quad \tan \alpha=\tan \frac{\pi}{3}-\frac{\pi}{4} \\ & \Rightarrow \quad \tan \alpha=\tan \frac{\pi}{12} \\ & \therefore \quad \alpha=\frac{\pi}{12} \end{aligned} $

From Eq. (i), $r \sin \alpha \cos \theta+r \cos \alpha \sin \theta=2$

$ \begin{matrix} \Rightarrow & r[\sin (\theta+\alpha)] =2 \\ & \sin (\theta+\alpha) =\frac{2}{2 \sqrt{2}} \\ \Rightarrow & \sin (\theta+\alpha) =\frac{1}{\sqrt{2}} \\ \Rightarrow & \sin (\theta+\alpha) =\sin \frac{\pi}{4} \theta+\alpha=n \pi+(-1)^{n} \frac{\pi}{4} \\ & \theta =n \pi+(-1)^{n} \cdot \frac{\pi}{4}-\frac{\pi}{12} \end{matrix} $

Alternate Method

$\begin{aligned} (\sqrt{3}-1) \cos \theta+(\sqrt{3}+1) \sin \theta & =2 \\ \text{Put}\quad \sqrt{3}-1 & =r \cos \alpha \text { and } \sqrt{3}+1=r \sin \alpha \\ \therefore \quad r & =2 \sqrt{2} \\ \end{aligned}$

Now,$\quad \tan \alpha=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$

$\begin{aligned} &\Rightarrow \tan \alpha=\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4} \cdot \tan \frac{\pi}{6}} \\ &\Rightarrow \tan \alpha=\tan \frac{\pi}{4}+\frac{\pi}{6} \Rightarrow \tan \alpha=\tan \frac{5 \pi}{12} \end{aligned}$

$\alpha=\frac{5 \pi}{12}$

From Eq. (i), $r \cos \alpha \cos \theta + r \sin \alpha \sin \theta =2$

$\qquad r[\cos(\theta - \alpha)]$

$\Rightarrow \qquad \cos(\theta - \alpha) = \frac{2}{2 \sqrt{2}} $

$\Rightarrow \qquad \cos(\theta - \alpha) = \frac{1}{ \sqrt{2}} $

$ \Rightarrow \qquad \cos(\theta - \alpha) = \cos \frac{\pi}{4}$

$\Rightarrow \qquad \theta - \alpha = 2n \pi \pm \frac{\pi}{4}$

$\therefore \qquad \theta = 2n \pi \pm \frac{\pi}{4}+\frac{5 \pi}{12}$

Solution

(c) Given that, $\sin \theta+cosec \theta=2$

$\Rightarrow \quad \sin ^{2} \theta+cosec^{2} \theta+2 \sin \theta \cdot cosec \theta=4$

$\Rightarrow \quad \sin ^{2} \theta+cosec^{2} \theta=4-2$

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

Solution

(d) Given that, $f(x)=\cos ^{2} x+\sec ^{2} x$

We know that, $A M \geq G M$

$ \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \cdot \sec ^{2} x} $

$\Rightarrow$

$\cos ^{2} x+\sec ^{2} x \geq 2$

$[\because \cos x \cdot \sec x=1]$

$\Rightarrow \qquad f(x) \geq 2$

Solution

(d) Given that,

$ \begin{matrix} \text { Now, } & \tan (\theta+\varphi)=\frac{\tan \theta+\tan \varphi}{1-\tan \theta \cdot \tan \varphi} \\ \Rightarrow & \tan (\theta+\varphi)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}} \Rightarrow \tan (\theta+\varphi)=\frac{\frac{3+2}{\frac{6}{6}}}{\frac{6-1}{6}}=\frac{5}{5}=1 \\ \Rightarrow & \tan (\theta+\varphi)=\tan \frac{\pi}{4} \\ \therefore & \theta+\varphi=\frac{\pi}{4} \end{matrix} $

Solution

(c) We know that, the range of $\sec \theta$ is $R-(-1,1)$.

Hence, $\sec \theta$ cannot be equal to $\frac{1}{2}$.

Solution

(b) Given expression, $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$

$ \begin{aligned} & =\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 45^{\circ} \cdot \tan (90^{\circ}-44^{\circ}) \tan (90^{\circ}-43^{\circ}) \ldots \tan (90^{\circ}-1^{\circ}) \\ & =\tan 1^{\circ} \cdot \cot 1^{\circ} \cdot \tan 2^{\circ} \cdot \cot 2^{\circ} \ldots \tan 89^{\circ} \cdot \cot 89^{\circ} \\ & =1 \cdot 1 \ldots 1 \cdot 1=1 \end{aligned} $

Solution

(c) Given expression, $\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}$

$ \begin{aligned} & \text { Let } \theta=15^{\circ} \\ & \text { We know that, } \cos 2 \theta=\overline{1+\tan ^{2} \theta} \\ & \therefore \quad \cos 30^{\circ}=\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}} \\ & \Rightarrow \quad \frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}=\frac{\sqrt{3}}{2} \quad [\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}] \end{aligned} $

Solution

(b) Given expression, $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ}$

$ \begin{aligned} & =\cos 1^{\circ} \cos 2^{\circ} \ldots \cos 90^{\circ} \ldots \cos 179^{\circ} \quad[\because \cos 90^{\circ}=0] \\ & =0 \end{aligned} $

Solution

(c)

$ \text {Given that,} \tan \theta=3 $

$ \Rightarrow \qquad \sec ^{2} \theta=1+\tan ^{2} \theta $

$\Rightarrow \qquad \sec \theta=\sqrt{1+9}= \pm \sqrt{10}$

$\Rightarrow \qquad \sec \theta=-\sqrt{10}$

$\Rightarrow \qquad \cos \theta=-\frac{1}{\sqrt{10}}$

$\Rightarrow \quad \sin \theta= \pm \sqrt{1-\frac{1}{10}}= \pm \sqrt{\frac{9}{10}}= \pm \frac{3}{\sqrt{10}}$ [since, $\theta$ lies in third quadrant]

$\therefore \quad \sin \theta=-\frac{3}{\sqrt{10}}$

Solution

(a) Given expression, $\tan 75^{\circ}-\cot 75^{\circ}$

$ \begin{aligned} & =\frac{\sin 75^{\circ}}{\cos 75^{\circ}}-\frac{\cos 75^{\circ}}{\sin 75^{\circ}} \\ & =\frac{\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ}}{\sin 75^{\circ} \cdot \cos _7{ }^{\circ}} \\ & =\frac{-2 \cos 150^{\circ}}{\sin 150^{\circ}} \\ & =\frac{-2 \cos (90^{\circ}+60^{\circ})}{\sin (90^{\circ}+60^{\circ})} \\ & =\frac{+2 \sin 60^{\circ}}{\cos 60^{\circ}} \\ & =\frac{2 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}}=2 \sqrt{3} \end{aligned} $

Solution

(b) We know that, if $\theta$ is increasing, then $\sin \theta$ is also increasing.

$\therefore \quad \sin 1^{\circ}<\sin 1 \qquad [\because 1 rad = 57^{\circ 30}]$

Solution

(d) Given that, $\tan \alpha=\frac{m}{m+1}$ and $\tan \beta=\frac{1}{2 m+1}$

Now,$\qquad \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$

$ \begin{matrix} \Rightarrow & \tan (\alpha+\beta)=\frac{\frac{m}{m+1}+\frac{1}{2 m+1}}{1-\frac{m}{m+1} \frac{1}{2 m+1}} \\ \\ \Rightarrow & \tan (\alpha+\beta)=\frac{m(2 m+1)+m+1}{(m+1)(2 m+1)-m} \\ \\ \Rightarrow & \tan (\alpha+\beta)=\frac{2 m^{2}+m+m+1}{2 m^{2}+2 m+m+1-m} \\ \\ \Rightarrow & \tan (\alpha+\beta)=\frac{2 m^{2}+2 m+1}{2 m^{2}+2 m+1} \Rightarrow \tan (\alpha+\beta)=1 \\ \\ \therefore & \alpha+\beta=\frac{\pi}{4} \end{matrix} $

Solution

(d) Given expression, $3 \cos x+4 \sin x+8$

$ \begin{matrix} \text { Let } & y =3 \cos x+4 \sin x+8 \\ \Rightarrow & y-8 =3 \cos x+4 \sin x \\ \therefore & \text { Minimum value of } y-8 =-\sqrt{9+16} \\ \Rightarrow & y-8 =-5 \Rightarrow y=-5+8 \\ \therefore & y =3 \end{matrix} $

Hence, the minimum value of $3 \cos x+4 \sin x+8$ is 3 .

Solution

(a)

$ \begin{aligned} \text {Let} \qquad 3 A & =A+2 A \\ \tan 3 A & =\tan (A+2 A) \end{aligned} $

$ \Rightarrow \quad \tan 3 A=\frac{\tan A+\tan 2 A}{1-\tan A \cdot \tan 2 A} $

$\Rightarrow \quad \tan A+\tan 2 A=\tan 3 A-\tan 3 A \cdot \tan 2 A \cdot \tan A$

$\Rightarrow \quad \tan 3 A-\tan 2 A-\tan A=\tan 3 A \cdot \tan 2 A \cdot \tan A$

Solution

(d) Given expression,

$ \begin{aligned} \sin (45^{\circ}+\theta) & -\cos (45^{\circ}-\theta) \\ & =\sin 45^{\circ} \cdot \cos \theta+\cos 45^{\circ} \cdot \sin \theta-\cos 45^{\circ} \cdot \cos \theta-\sin 45^{\circ} \cdot \sin \theta \\ & =\frac{1}{\sqrt{2}} \cdot \cos \theta+\frac{1}{\sqrt{2}} \cdot \sin \theta-\frac{1}{\sqrt{2}} \cdot \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \\ & =0 \end{aligned} $

Solution

(c)

$ \begin{aligned} &\text {Given expression,} \qquad (\cot \frac{\pi}{4}+\theta)-\cot (\frac{\pi}{4}-\theta) \\ & =(\frac{\cot \frac{\pi}{4} \cot \theta-1}{\cot \frac{\pi}{4}+\cot \theta}) \cdot (\frac{\cot \frac{\pi}{4} \cot \theta+1}{\cot \theta-\cot \frac{\pi}{4}}) \\ & =(\frac{\cot \theta-1}{\cot \theta+1}) \cdot (\frac{\cot \theta+1}{\cot \theta-1}) \\ & =1 \end{aligned} $

Solution

(b) Given expression, $\cos 2 \theta \cos 2 \varphi+\sin ^{2}(\theta-\varphi)-\sin ^{2}(\theta+\varphi)$

$ \begin{aligned} & =\cos 2 \theta \cdot \cos 2 \varphi+\sin (\theta-\varphi+\theta+\varphi) \cdot \sin (\theta-\varphi-\theta-\varphi) \\ & =\cos 2 \theta \cdot \cos 2 \varphi-\sin 2 \theta \cdot \sin 2 \varphi \\ & =\cos (2 \theta+2 \varphi)=\cos 2(\theta+\varphi) \end{aligned} $

Solution

(c) Given expression, $\cos 12^{\circ}+\cos 84^{\circ}+\cos 150^{\circ}+\cos 132^{\circ}$

$=\cos 12^{\circ}+\cos 150^{\circ}+\cos 84^{\circ}+\cos 132^{\circ}$

$=2 \cos (\frac{12^{\circ}+150^{\circ}}{2}) \cdot \cos (\frac{12^{\circ}-150^{\circ}}{2})+2 \cos (\frac{84^{\circ}+132^{\circ}}{2}) \cdot \cos (\frac{84^{\circ}-132^{\circ}}{2})$

$=2 \cos 84^{\circ} \cos 72^{\circ}+2 \cos 108^{\circ} \cdot \cos 24^{\circ}$

$=2 \cos 84^{\circ} \cos (90^{\circ}-18^{\circ})+2 \cos (90^{\circ}+18^{\circ}) \cdot \cos 24^{\circ}$

$=2 \cos 84^{\circ} \sin 18^{\circ}-2 \sin 18^{\circ} \cdot \cos 24^{\circ}$

$=2 \sin 18^{\circ}(\cos 84^{\circ}-\cos 24^{\circ})$

$=2 \sin 18^{\circ} \cdot 2 \sin (\frac{84^{\circ}+24^{\circ}}{2} )\cdot \sin (\frac{84^{\circ}-24^{\circ}}{2})$

$=-4 \sin 18^{\circ} \cdot \sin 54^{\circ} \sin 30^{\circ}$

$=-4 \frac{\sqrt{5}-1}{4} \cdot \cos 36^{\circ} \cdot \frac{1}{2}$

$=-(\sqrt{5}-1) \frac{\sqrt{5}+1}{4} \cdot \frac{1}{2}=-\frac{5-1}{8}=\frac{-4}{8}=\frac{-1}{2}$

Solution

(c) Given that, $\tan A=\frac{1}{2}$ and $\tan B=\frac{1}{3}$

Now, $\quad \tan (2 A+B)=\frac{\tan 2 A+\tan B}{1-\tan 2 A \cdot \tan B}$

Also, $\quad \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}=\frac{2 \cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}$

From Eq. (i), $\quad \tan (2 A+B)=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3} \cdot \frac{1}{3}}=\frac{\frac{4}{3}+\frac{1}{3}}{\frac{9-4}{9}}=\frac{\frac{5}{3}}{\frac{5}{9}}=3$

Solution

(c) Given expression, $\sin \frac{\pi}{10} \sin \frac{13 \pi}{10}=\sin \frac{\pi}{10} \sin \pi+\frac{3 \pi}{10}$

$ \begin{aligned} & =-\sin \frac{\pi}{10} \sin \frac{3 \pi}{10}=-\sin 18^{\circ} \cdot \sin 54^{\circ} \\ & =-\sin 18^{\circ} \cdot \cos 36^{\circ} \\ & =-\frac{\sqrt{5}-1}{4} \frac{\sqrt{5}+1}{4} \quad [\text{since,put this value here}] \\ & =-\frac{5-1}{16}=-\frac{1}{4} \end{aligned} $

Solution

(b) Given expression, $\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$

$ \begin{aligned} & =2 \cos (\frac{50^{\circ}+70^{\circ}}{2}) \cdot \sin (\frac{50^{\circ}-70^{\circ}}{2}+\sin 10^{\circ}) \\ & =-2 \cos 60^{\circ} \sin 10^{\circ}+\sin 10^{\circ} \\ & =-2 \cdot \frac{1}{2} \sin 10^{\circ}+\sin 10^{\circ}=0 \end{aligned} $

Solution

(c) Given that, $\sin \theta+\cos \theta=1$

On squaring both sides, we get

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

$\Rightarrow \qquad 1+\sin 2 \theta =1 $

$\therefore \qquad \sin 2 \theta =0 $

Solution

(b) Given that, $\alpha+\beta=\frac{\pi}{4}$

Now, $\quad(1+\tan \alpha)(1+\tan \beta)=1+\tan \alpha+\tan \beta+\tan \alpha \tan \beta \quad …(i)$

We know that, $\quad \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$

$ \begin{aligned} & \Rightarrow \quad 1=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \quad ..(i) \\ & \Rightarrow \quad \tan \alpha+\tan \beta=1-\tan \alpha \tan \beta \end{aligned} $

From Eq. (i),

$ \begin{aligned} (1+\tan \alpha)(1+\tan \beta) & =1+1-\tan \alpha \cdot \tan \beta+\tan \alpha \cdot \tan \beta \\ & =2 \end{aligned} $

Solution

(c) Given that, $\quad \sin \theta=\frac{-4}{5}$

$ \begin{aligned} \cos \theta & =\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-16}{25}}= \pm \frac{3}{5} \\ \cos \theta & =\frac{-3}{5} \quad \text { [since, } \theta \text { lies in third quadrant] }\\ \Rightarrow \quad 2 \cos ^{2} \frac{\theta}{2}-1 & =\frac{-3}{5} \\ \Rightarrow \quad 2 \cos ^{2} \frac{\theta}{2} & =1-\frac{3}{5} \\ \Rightarrow \quad 2 \cos ^{2} \frac{\theta}{2} & =\frac{2}{5} \\ \therefore \quad \cos ^{\frac{\theta}{2}} & = \pm \frac{1}{\sqrt{5}} \\ \Rightarrow \quad \cos \frac{\theta}{2} & =-\frac{1}{\sqrt{5}} \quad \text { [since, } \theta \text { lies in third quadrant] } \end{aligned} $

Solution

(c) Given equation,

$\Rightarrow \qquad \tan +\sec x=2 \cos x$

$\Rightarrow \qquad \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x$

$ \Rightarrow \qquad 1+\sin x=2 \cos ^{2} x$

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

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

$\Rightarrow \quad 2 \sin ^{2} x+\sin x-1=0$

$\Rightarrow \quad 2 \sin ^{2} x+2 \sin x-\sin x-1=0$

$\Rightarrow \quad 2 \sin x(\sin x+1)-1(\sin x+1)=0$

$\Rightarrow \quad(\sin x+1)(2 \sin x-1)=0$

$\Rightarrow \quad \sin x+1=0$ or $(2 \sin x-1)=0$

$\Rightarrow \quad \sin x=-1, \sin x=\frac{1}{2}$

$\therefore \quad x=\frac{3 \pi}{2}, x=\frac{\pi}{6}$

Hence, only two solutions possible.

Solution

(a) Given expression, $\sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18}$

$ \begin{aligned} & =\sin 10^{\circ}+\sin 20^{\circ}+\sin 40^{\circ}+\sin 50^{\circ} \\ & =\sin 50^{\circ}+\sin 10^{\circ}+\sin 40^{\circ}+\sin 20^{\circ} \\ & =\sin 130^{\circ}+\sin 10^{\circ}+\sin 140^{\circ}+\sin 20^{\circ} \\ & =2 \sin 70^{\circ} \cos 60^{\circ}+2 \sin 80^{\circ} \cdot \cos 60^{\circ} \quad [\because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} ]\\ & =2 \cdot \frac{1}{2} \sin 70^{\circ}+2 \cdot \frac{1}{2} \sin 80^{\circ} \quad [\because \cos 60^{\circ}=\frac{1}{2}] \\ & =\sin 70^{\circ}+\sin 80^{\circ}=\sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9} \end{aligned} $

Solution (b)

$ \begin{aligned} & \text{Given equation,} \qquad 3 \tan A+4=0 \\ & \Rightarrow \qquad 3 \tan A=-4 \\ & \Rightarrow \qquad \tan A=\frac{-4}{3} \\ &\Rightarrow \qquad \cot A=\frac{-3}{4} \\ &\Rightarrow \qquad \sec A=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}= \pm \frac{5}{3} \quad \text { [since, A lies in second quadrant] } \\ &\Rightarrow \qquad \sec A=\frac{-5}{3} \\ & \cos A=\frac{-3}{5} \\ & \sin A=\sqrt{1-\frac{9}{25}}=\frac{\sqrt{25-9}}{25}= \pm \frac{4}{5} \\ & \sin A=\frac{4}{5} \end{aligned} \quad \text { [since, A lies in second quadrant] } $

$ \begin{aligned} \therefore \quad 2 \cot A-5 \cos A+\sin A & =2 \frac{-3}{4}-5 \frac{-3}{5}+\frac{4}{5} \\ & =\frac{-6}{4}+3+\frac{4}{5} \\ & =\frac{-30+60+16}{20}=\frac{46}{20} \\ & =\frac{23}{10} \end{aligned} $

Solution

(a) Given expression, $\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$

$ \begin{aligned} & =\cos (48^{\circ}+12^{\circ})-\cos (48^{\circ}-12^{\circ}) \\ & =\cos 60^{\circ} \cdot \cos 36^{\circ} \\ & =\frac{1}{2} \cdot \frac{\sqrt{5}+1}{4} \\ & =\frac{\sqrt{5}+1}{8} \end{aligned} $

Solution

(b) Given that,

$ \begin{aligned} \cos 2 \alpha & =\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{\frac{48}{49}}{\frac{50}{49}} \\ & =\frac{48}{50}=\frac{24}{25} \\ \Rightarrow \quad \cos 2 \alpha & =\frac{24}{25} \quad ..(i) \\ \text { We know that, } \quad \sin 4 \beta & =\frac{2 \tan 2 \beta}{1+\tan ^{2} 2 \beta} ..(ii) \\ \text { and } \quad \tan 2 \beta & =\frac{2 \tan ^{2}}{1-\tan ^{2} \beta}=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}} \\ & =\frac{2}{\frac{3}{9}}=\frac{2 \times 9}{3 \times 8}=\frac{3}{4} \end{aligned} $

$ \tan =\frac{1}{7} \text { and } \tan \beta=\frac{1}{3} $

From Eq, (ii),

$\sin 4 \beta=\frac{2 \times \frac{3}{4}}{1+\frac{9}{16}}=\frac{\frac{6}{4}}{\frac{25}{16}}=\frac{6 \times 16}{4 \times 25} $

$\Rightarrow \qquad \sin 4 \beta=\frac{24}{25} $

$\Rightarrow \qquad \sin 4 \beta=\cos 2 \alpha$

$\therefore \qquad \cos 2 \alpha=\sin 4 \beta$

Solution

(b) Given that, $\tan \theta=\frac{a}{b}$

$ \begin{aligned} \therefore \quad b \cos 2 \theta+a \sin 2 \theta & =b (\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta})+a (\frac{2 \tan \theta}{1+\tan ^{2} \theta}) \\ & =b (\frac{1-\frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}})+a( \frac{\frac{2 a}{b}}{1+\frac{a^{2}}{b^{2}}}) \\ & =b (\frac{b^{2}-a^{2}}{b^{2}+a^{2}})+\frac{2 a^{2} b}{a^{2}+b^{2}} \\ & =\frac{b}{a^{2}+b^{2}}[b^{2}-a^{2}+2 a^{2}]=\frac{(a^{2}+b^{2}) b}{(a^{2}+b^{2})} \\ & =b \end{aligned} $

Solution

(d)

$ \begin{aligned} \text{Here,} \cos \theta & =x+\frac{1}{x} \\ \Rightarrow \qquad \cos \theta & =\frac{x^{2}+1}{x} \\ x^{2}-x \cos \theta+1 & =0 \end{aligned} $

For real value of $x, \quad(-\cos \theta)^{2}-4 \times 1 \times 1=0$

$ \begin{aligned} \cos ^{2} \theta & =4 \\ \cos \theta & = \pm 2 \end{aligned} $

which is not possible.

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

Solution

Here,

$ \begin{aligned} \frac{\sin 50^{\circ}}{\sin 130^{\circ}} & =\frac{\sin (180^{\circ}-130^{\circ})}{\sin 130^{\circ}} \\ & =\frac{\sin 130^{\circ}}{\sin 130^{\circ}}=1 \end{aligned} $

Solution Here, $\quad k=\sin (\frac{\pi}{18}) \sin (\frac{5 \pi}{18}) \sin (\frac{7 \pi}{18})$

$ =\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} $

$ \begin{aligned} & =\sin 10^{\circ} \cos 40^{\circ} \cdot \cos 20^{\circ} \\ & =\frac{1}{2} \sin 10^{\circ}[2 \cos 40^{\circ} \cdot \cos 20^{\circ}] \\ & =\frac{1}{2} \sin 10^{\circ}[\cos 60^{\circ}+\cos 20^{\circ}] \quad[\because 2 \cos x \cdot \cos y=\cos (x+y)+\cos (x-y)] \\ & =\frac{1}{2} \sin 10^{\circ} \cdot \frac{1}{2}+\frac{1}{2} \sin 10^{\circ} \cos 20^{\circ} \\ & =\frac{1}{4} \sin 10^{\circ}+\frac{1}{4}[\sin 30^{\circ}-\sin 10^{\circ}] \\ & =\frac{1}{8} \end{aligned} $

Solution

Given that,

$ \begin{aligned} \tan A & =\frac{1-\cos B}{\sin B} \\ & =\frac{1-1+2 \sin ^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}}=\tan \frac{B}{2} \end{aligned} $

$\text{Now,} \qquad \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$

$\Rightarrow \qquad \tan 2 A=\frac{2 \cdot \tan \frac{B}{2}}{1-\tan ^{2} \frac{B}{2}}$

$\Rightarrow \qquad \tan 2 A=\tan B$

Solution

Given that, $\sin x+\cos x=a$

On squaring both sides, we get

$ \begin{aligned} & (\sin x+\cos x)^{2}=(a)^{2} \\ & \Rightarrow \quad \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2} \\ & \Rightarrow \quad \sin x \cdot \cos x=\frac{1}{2}(a^{2}-1) \\ (i)& \sin^{6}x + \cos^{6}x=(sin^{2}x)^{3} + (\cos^{2}x)^3\\ & =(\sin ^{2} x+\cos ^{2} x)(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x) \\ & =\sin ^{4} x+\cos ^{4} x-\frac{1}{4}(a^{2}-1)^{2} \\ & =(\sin ^{2} x+\cos ^{2} x)^{2}-2 \sin ^{2} x \cos ^{2} x-\frac{1}{4}(a^{2}-1)^{2} \\ & =1-2 \cdot \frac{1}{4}(a^{2}-1)^{2}-\frac{1}{4}(a^{2}-1)^{2}=\frac{1}{4}[4-3(a^{2}-1)^{2}] \end{aligned} $

(ii) $|\sin x-\cos x|=\sqrt{(\sin x-\cos x)^{2}}$

$ \begin{aligned} & =\sqrt{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x} \\ & =\sqrt{1-2 \frac{1}{2}(a^{2}-1)}=\sqrt{1-a^{2}+1}=\sqrt{2-a^{2}} \end{aligned} $

Solution

In right angled $\triangle A B C, \angle C=90^{\circ}$

$ \begin{aligned} & \therefore \quad \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ & \Rightarrow \quad \frac{1}{0}=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ & \Rightarrow \quad \tan A \tan B=1 \\ & \tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B} \\ & =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \\ & =\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cdot \cos A} \\ & =\frac{1}{\sin A \cdot \cos A}=\frac{2}{2 \cdot \sin A \cdot \cos A} \end{aligned} $

$ \begin{matrix} =\frac{\sin A}{\cos A}+\frac{\sin (90^{\circ}-A)}{\cos (90^{\circ}-A)} & {[\because \angle C=90^{\circ}, \angle B=90^{\circ}-A]} \\ =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} & \\ =\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cdot \cos A} & \\ =\frac{1}{\sin A \cdot \cos A}=\frac{2}{2 \cdot \sin A \cdot \cos A} & \\ =\frac{2}{\sin 2 A} & {[\because \sin 2 x=2 \sin x \cos x]} \end{matrix} $

So, the required equation is $x^{2}-\frac{2}{\sin A} x+1$.

$\because \sin 2 x=2 \sin x \cos x$

Solution

Given expression, $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$

$ \begin{aligned} & =3[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x]^{2}+6[\sin ^{2} x+\cos ^{2} x+2 \cdot \sin x \cdot \cos x] \\ & \quad+4[(\sin ^{2} x)^{3}+(\cos ^{2} x)^{3}] \\ & =3(1-\sin 2 x)^{2}+6(1+\sin 2 x)+4[(\sin ^{2}+\cos ^{2} x)(\sin ^{4} x-\sin x \cos ^{2} x+\cos ^{4} x). \\ & =3(1+\sin ^{2} 23 x-2 \sin 2 x)+6+6 \sin 2 x+4[(\sin ^{2} x+\cos ^{2} x)^{2} 3 \sin x \cos ^{2} x] \\ & =3+3 \sin ^{2} 2 x-6 \sin 2 x+6+6 \sin 2 x \\ & =4-3 \sin ^{2} 2 x=13 \end{aligned} $

Solution

Given function, $f(x)=-3 \cos \sqrt{3+x+x^{2}}$

$ \begin{matrix} \text{We know that,}& -1 \leq \cos x \leq 1 \\ \Rightarrow & -3 \leq 3 \cos x \leq 3 \\ \Rightarrow & 3 \geq-3 \cos x \geq- \\ \Rightarrow & -3 \leq-3 \cos x \leq 3 \end{matrix} $

So, the value of $f(x)$ lies in $[-3,3]$.

Solution

Given that, $y=\sqrt{3} \sin x+\cos x$

$ \begin{aligned} y & =2 [\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x ]\\ & =2 [\sin x \cdot \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6} ]\\ & =2 \sin (x+\pi / 6) \end{aligned} $

Graph of $y=2 \sin x$

Hence, the maximum distance is 2 units.

Solution

True

$ \begin{aligned} \text{Given that,} & \tan A=\frac{1-\cos B}{\sin B}=\frac{1-1+2 \sin ^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}}=\tan \frac{B}{2} \\ \text{Now,} & \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}=\frac{2 \cdot \tan \frac{B}{2}}{1-\tan ^{2} \frac{B}{2}}=\tan B \end{aligned} $

Solution

False

Given that, $\sin A+\sin 2 A+\sin 3 A=3$

It is possible only if $\sin A, \sin 2 A, \sin 3 A$ each has a value one because maximum value of $\sin A$ is a certain angle is 1 . Which is not possible because angle are different.

Solution

False

$\qquad \sin 10^{\circ} =\sin (90^{\circ}-80^{\circ}) $

$\qquad \sin 10^{\circ} =\cos 80^{\circ} $

$\because \qquad \cos 80^{\circ} <\cos 10^{\circ} $

$\text{Hence,}\qquad \sin 10^{\circ} <\cos 10^{\circ}$

Solution

True

$ \begin{aligned} \text { LHS } & =\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15} \\ & =\cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \cos 192^{\circ} \\ & =\frac{1}{16 \sin 24^{\circ}}[(2 \sin 24^{\circ} \cos 24^{\circ})(2 \cos 48^{\circ})(2 \cos 96^{\circ})(2 \cos 192^{\circ})] \\ & =\frac{1}{16 \sin 24^{\circ}}[2 \sin 48^{\circ} \cos 48^{\circ}(2 \cos 96^{\circ})(2 \cos 192^{\circ})] \\ & =\frac{1}{16 \sin 24^{\circ}}[(2 \sin 96^{\circ} \cos 96^{\circ})(2 \cos 192^{\circ})] \\ & =\frac{1}{16 \sin 24^{\circ}}[2 \sin 192^{\circ} \cos 192^{\circ}) \\ & =\frac{1}{16 \sin 24^{\circ}} \sin 384^{\circ}=\frac{\sin (360^{\circ}+24^{\circ})}{16 \sin 24^{\circ}} \\ & =\frac{1}{16}=RHS \qquad \text { Hence proved. } \end{aligned} $

Solution

False

Given equation, $\quad \sin ^{4} \theta-2 \sin ^{2} \theta-1=0$

$\Rightarrow \qquad \sin ^{2} \theta=\frac{2 \pm \sqrt{4+4}}{2}$

$\Rightarrow \quad \sin ^{2} \theta=\frac{2 \pm 2 \sqrt{2}}{2}$

$\Rightarrow \quad \sin ^{2} \theta=(1+\sqrt{2})$ or $(1-\sqrt{2}) \Rightarrow-1 \leq \sin \theta \leq 1$

$\Rightarrow \quad \sin ^{2} \theta \leq 1$

$\because \quad \sin ^{2} \theta=\sqrt{2+1}$ or $(1-\sqrt{2})$

which is not possible.

Solution

True

Given that,

$ \begin{aligned} cosec x & =1+\cot x \\ \frac{1}{\sin x} & =1+\frac{\cos x}{\sin x} \Rightarrow \frac{1}{\sin x}=\frac{\sin x+\cos x}{\sin x} \end{aligned} $

$\Rightarrow$

$\Rightarrow \quad \sin x+\cos x=1$

$\Rightarrow \quad \frac{1}{\sqrt{2}} \cdot \sin x+\frac{1}{\sqrt{2}} \cdot \cos x=\frac{1}{\sqrt{2}}$

$\Rightarrow \quad \sin \frac{\pi}{4} \sin x+\cos x \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

$\Rightarrow \quad \cos( x-\frac{\pi}{4})=\cos \frac{\pi}{4}$

$ \therefore \quad x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4} $

$ \begin{aligned} \text{For positive sign,} & x=2 n \pi+\frac{\pi}{4}+\frac{\pi}{4}=2 n \pi+\frac{\pi}{2} \\ \text{For negative sign,} & x=2 n \pi-\frac{\pi}{4}+\frac{\pi}{4}=2 n \pi \end{aligned} $

Solution

True

$ \begin{aligned} & \tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3} \\ & \Rightarrow \quad \tan \theta+\tan 2 \theta=\sqrt{3}-\sqrt{3} \tan \theta \tan 2 \theta \\ & \Rightarrow \quad \tan \theta+\tan 2 \theta=\sqrt{3}(1-\tan \theta \tan 2 \theta) \\ & \Rightarrow \quad \frac{\tan \theta+\tan 2 \theta}{1-\tan \theta \tan 2 \theta}=\sqrt{3} \\ & \Rightarrow \quad \tan (\theta+2 \theta)=\tan \frac{\pi}{3} \Rightarrow \tan 3 \theta=\tan \frac{\pi}{3} \\ & \therefore \quad 3 \theta=n \pi+\frac{\pi}{3} \\ & \theta=\frac{n \pi}{3}+\frac{\pi}{9} \end{aligned} $

Solution

True

$ \text{We have,} \qquad \tan (\pi \cos \theta)=\cot (\pi \sin \theta) $

$ \Rightarrow \qquad \tan (\pi \cos \theta)=\tan \frac{\pi}{2}-(\pi \sin \theta) $

$\Rightarrow \qquad \pi \cos =\frac{\pi}{2}-\pi \sin \theta$

$\Rightarrow \quad \pi(\sin \theta+\cos \theta)=\frac{\pi}{2}$

$\Rightarrow \quad \sin \theta+\cos \theta=\frac{1}{2}$

$\Rightarrow \quad \frac{1}{\sqrt{2}} \cdot \sin \theta+\frac{1}{\sqrt{2}} \cdot \cos \theta=\frac{1}{2 \sqrt{2}}$

$\Rightarrow \quad \sin \theta \cdot \sin \frac{\pi}{4}+\cos \theta \cdot \cos \frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$

$\therefore \quad \cos \theta-\frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$

Solution

(i) $\sin (x+y) \sin (x-y)=\sin ^{2} x-\sin ^{2} y$

(ii) $\cos (x+y) \cos (x-y)=\cos ^{2} x-\sin ^{2} y$

(iii) $\cot \frac{\pi}{4}+\theta=\frac{\cot \frac{\pi}{4} \cot \theta-1}{\cot \frac{\pi}{4}+\cot \theta}$

$=\frac{-1+\cot \theta}{1+\cot \theta}=\frac{1-\tan \theta}{1+\tan \theta}$

(iv) $\tan \frac{\pi}{4}+\theta=\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}=\frac{1+\tan \theta}{1-\tan \theta}$

Hence, the correct mathes are (i) $\rarr$ (d), (ii) $\rarr$ (a), (iii) $\rarr$ (b), (iv) $\rarr$ (c).