Maths Solve the Following Question.(3 Marks)

$\begin{aligned} & \cot \theta=0=\cot \frac{\pi}{2}=\cot \left(\pi+\frac{\pi}{2} \ldots[\because \cos (\pi+\theta)=\cot \theta]\right. \\ & \therefore \cot \theta=\cot \frac{\pi}{2}=\cot \frac{3 \pi}{2}\end{aligned}$

$\therefore \theta=\frac{\pi}{2}$ or $\theta=\frac{3 \pi}{2}$

Hence, the required principal solutions are $\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}$

$\begin{aligned} \therefore \tan 3 \theta & =-1=-\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right) \\ & =\tan \left(2 \pi-\frac{\pi}{4}\right)=\tan \left(3 \pi-\frac{\pi}{4}\right) \\ & =\tan \left(4 \pi-\frac{\pi}{4}\right)=\tan \left(5 \pi-\frac{\pi}{4}\right) \\ & =\tan \left(6 \pi-\frac{\pi}{4}\right)\end{aligned}$

$\ldots[\because \tan (\pi-\theta)=\tan (2 \pi-\theta)=\tan (3 \pi-\theta)$

$\begin{aligned} & =\tan (4 \pi-\theta)=\tan (5 \pi-\theta)=\tan (6 \pi-\theta)=-\tan \theta] \\ & \therefore \tan 3 \theta=\tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}=\tan \frac{11 \pi}{4}=\tan \frac{15 \pi}{4}\end{aligned}$

$=\tan \frac{19 \pi}{4}=\tan \frac{23 \pi}{4}$

$\therefore 3 \theta=\frac{3 \pi}{4}$ or $3 \theta=\frac{7 \pi}{4}$ or $3 \theta=\frac{11 \pi}{4}$ or $3 \theta=\frac{15 \pi}{4}$

or $3 \theta=\frac{19 \pi}{4}$ or $3 \theta=\frac{23 \pi}{4}$

$\therefore \theta=\frac{\pi}{4}$ or $\theta=\frac{7 \pi}{12}$ or $\theta=\frac{11 \pi}{12}$ or $\theta=\frac{5 \pi}{4}$

or $\theta=\frac{19 \pi}{12}$ or $\theta=\frac{23 \pi}{12}$

$\left\{\frac{\pi}{4}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{5 \pi}{4}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\right\}$.

Hence, the required principal solutions are,

$\left\{\frac{\pi}{4}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{5 \pi}{4}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\right\}$.

Since, $\theta \in(0,2 \pi), 2 \in \in(0,4 \pi)$

$\therefore \sin 2 \theta=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(2 \pi-\frac{\pi}{6}\right)$

$=\sin \left(3 \pi+\frac{\pi}{6}\right)=\sin \left(4 \pi-\frac{\pi}{6}\right)$

$\ldots[\because \sin (\pi+\theta)=\sin (2 \pi-\theta)=\sin (3 \pi+\theta)$

$=\sin (4 \pi-\theta)=-\sin \theta]$

$\therefore \sin 2 \theta=\sin \frac{7 \pi}{6}=\sin \frac{11 \pi}{6}=\sin \frac{19 \pi}{6}=\sin \frac{23 \pi}{6}$

$\therefore 2 \theta=\frac{7 \pi}{6}$ or $2 \theta=\frac{11 \pi}{6}$ or $2 \theta=\frac{19 \pi}{6}$ or $2 \theta=\frac{23 \pi}{6}$

$\therefore \theta=\frac{7 \pi}{12}$ or $\theta=\frac{11 \pi}{12}$ or $\theta=\frac{19 \pi}{12}$ or $\theta=\frac{23 \pi}{12}$

Hence, the required principal solutions are

$\left\{\frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\right\}$.

$\begin{aligned} & \frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}=k \\ & \therefore \sin A=k a, \sin B=k b, \sin C=k c\end{aligned}$

Now, $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$

$\begin{aligned} & \therefore \sin A \cdot \sin (B-C)=\sin C \cdot \sin (A-B) \\ & \therefore \sin [\pi-(B+C)] \cdot \sin (B-C)\end{aligned}$

$\begin{aligned} & =\sin [\pi-(A+B)] \cdot \sin (A-B) \ldots[\because A+B+C=\pi] \\ & \therefore \sin (B+C) \cdot \sin (B-C)=\sin (A+B) \cdot \sin (A-B)\end{aligned}$

$\begin{aligned} & \therefore \sin ^2 B-\sin ^2 C=\sin ^2 A-\sin ^2 B \\ & \therefore 2 \sin ^2 B=\sin ^2 A+\sin ^2 C\end{aligned}$

$\begin{aligned} & \therefore 2 k^2 b^2=k^2 a^2+k^2 c^2 \\ & \therefore 2 b^2=a^2+c^2\end{aligned}$

Hence, $a^2, b^2, c^2$ are in A.P.

$\therefore 2 \cos \left(\frac{A+C}{2}\right) \cdot \cos \left(\frac{A-C}{2}\right)=\sin B$

$\therefore 2 \cos \left(\frac{\pi}{2}-\frac{B}{2}\right) \cdot \cos \left(\frac{A-C}{2}\right)=\sin B$

$\ldots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$

$\therefore 2 \sin \frac{B}{2} \cdot \cos \left(\frac{A-C}{2}\right)=2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}$

$\therefore \cos \left(\frac{A-C}{2}\right)=\cos \frac{B}{2}$

$\therefore \frac{A-C}{2}=\frac{B}{2}$

∴ A – C = B ∴ A = B + C ∴ A + B + C = 180° gives A + A = 180° ∴ 2A = 180 ∴ A = 90° ∴ ∆ ABC is a rightangled triangle.

$=\left(a^2+b^2-2 a b\right) \cos ^2 \frac{C}{2}+\left(a^2+b^2+2 a b\right) \sin \frac{\mathrm{C}}{2} 2$

$=\left(a^2+b^2\right) \cos ^2 \frac{\mathrm{C}}{2}-2 a b \cos ^2 \frac{\mathrm{C}}{2}+\left(a^2+b^2\right) \sin ^2 \frac{\mathrm{C}}{2}+2 a b \sin ^2 \frac{\mathrm{C}}{2}$

$=\left(a^2+b^2\right)\left(\cos ^2 \frac{\mathrm{C}}{2}+\sin ^2 \frac{\mathrm{C}}{2}\right)-2 a b\left(\cos ^2 \frac{\mathrm{C}}{2}-\sin ^2 \frac{\mathrm{C}}{2}\right)$

$\begin{aligned} & =a^2+b^2-2 a b \cos \mathrm{C} \\ & =c^2=\text { RHS. }\end{aligned}$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ $\therefore a=k \sin \mathrm{A}, b=k \sin \mathrm{B}, c=k \sin C$

$\begin{aligned} \text { RHS } & =\left(\frac{a+b}{c}\right) \sin \frac{\mathrm{C}}{2} \\ & =\left(\frac{k \sin \mathrm{A}+k \sin \mathrm{B}}{k \sin \mathrm{C}}\right) \sin \frac{\mathrm{C}}{2} \\ & =\left(\frac{\sin \mathrm{A}+\sin \mathrm{B}}{\sin \mathrm{C}}\right) \sin \frac{\mathrm{C}}{2}\end{aligned}$

$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \times \\ & 2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$

$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \times \\ & 2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$

Then, $\cos \alpha=x$ where $0<\alpha<\pi$

Since, $x<0$, $\frac{\pi}{2}<\alpha<\pi$

Now, $\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^2 \alpha}}{\cos \alpha}\right)$

$=\tan ^{-1}(\tan \alpha)$

$\ldots(1)$

But $\frac{\pi}{2}<\alpha<\pi$, therefore inverse of tangent does not exist.

Consider, $\frac{\pi}{2}-\pi<\alpha-\pi<\pi-\pi$

$\therefore \quad-\frac{\pi}{2}<\alpha-\pi<0$

and $\tan (\alpha-\pi)=\tan [-(\pi-\alpha)]$

$\begin{aligned} & =-\tan (\pi-\alpha) \\ & \quad \ldots[\because \tan (-\theta)=-\tan \theta]\end{aligned}$

$=-(-\tan \alpha)=\tan \alpha$

$\therefore$ from (1), we get

$\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=\tan ^{-1}[\tan (\alpha-\pi)]$

$\begin{aligned} & =\alpha-\pi \ldots\left[\because \tan ^{-1}(\tan x)=x\right] \\ & =\cos ^{-1} x-\pi\end{aligned}$

$\therefore \cos ^{-1} x=\pi+\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$, if $x<0$.

Then, $\cos \alpha=x$, where $0<\alpha<\pi$

Since, $x>0,0<\alpha<\frac{\pi}{2}$

$\therefore \sin \alpha>0, \cos \alpha>0$

Now, $\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^2 \alpha}}{\cos \alpha}\right)$

$\begin{aligned} & =\tan ^{-1}\left(\frac{\sqrt{\sin ^2 \alpha}}{\cos \alpha}\right) \\ & =\tan ^{-1}(\tan \alpha) \\ & =\alpha=\cos ^{-1} x\end{aligned}$

Hence, $\cos ^{-1} x=\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$, if $x>0$.

$\tan ^{-1} \frac{1}{2}=\frac{1}{3} \tan ^{-1} \frac{11}{2}$

i.e. to show that $3 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{11}{2}$

$\mathrm{LHS}^2 3 \tan ^{-1} \frac{1}{2}$

$=2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}$

$=\tan ^{-1}\left[\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right]+\tan ^{-1} \frac{1}{2}$

$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$

$=\tan ^{-1}\left[\frac{1}{(3 / 4)}\right]+\tan ^{-1} \frac{1}{2}$

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

$=\tan ^{-1}\left[\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3} \times \frac{1}{2}}\right]$

$\begin{aligned} & =\tan ^{-1}\left(\frac{8+3}{6-4}\right) \\ & =\tan ^{-1}\left(\frac{11}{2}\right)=\text { RHS. }\end{aligned}$

$=\tan ^{-1} 3-\tan ^{-1} \frac{1}{3} \quad \cdots\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$

$=\tan ^{-1}\left[\frac{3-\frac{1}{3}}{1+3\left(\frac{1}{3}\right)}\right]$

$=\tan ^{-1}\left[\frac{\left(\frac{8}{3}\right)}{1+1}\right]$

$=\tan ^{-1}\left(\frac{4}{3}\right)$

$\begin{aligned} & =\cot ^{-1}\left(\frac{3}{4}\right) \quad \cdots\left[\tan ^{-1} x=\cot ^{-1}\left(\frac{1}{x}\right)\right] \\ & =\text { RHS. }\end{aligned}$

$\therefore \tan ^{-1}\left[\frac{2 \cos x}{1-\cos ^2 x}\right]=\tan ^{-1}(2 \operatorname{cosec} x)$

$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$

$\therefore \frac{2 \cos x}{1-\cos ^2 x}=2 \operatorname{cosec} x$

$\begin{aligned} & \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \\ & \therefore \cos x=\sin x \\ & \therefore x=\frac{\pi}{4}\end{aligned}$

$\cdots\left[\because \sin \frac{\pi}{4}=\cos \frac{\pi}{4}\right]$

$\therefore \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4}$

$\therefore \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$

$\therefore \frac{\left(x^2+x-2\right)+\left(x^2-x-2\right)}{\left(x^2-4\right)-\left(x^2-1\right)}=1$

$\therefore \frac{x^2+x-2+x^2-x-2}{x^2-4-x^2+1}=1$

$\therefore \frac{2 x^2-4}{-3}=1$

$\begin{aligned} & \therefore 2 x^2-4=-3 \\ & \therefore 2 x^2=1 \quad \therefore x^2=\frac{1}{2} \\ & \therefore x= \pm \frac{1}{\sqrt{2}} .\end{aligned}$

$\begin{aligned} & \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=k \\ & \text { LHS }=a^2\left(\cos ^2 B-\cos ^2 C\right)+b^2\left(\cos ^2 C-\cos ^2 A\right)+c^2\left(\cos ^2 A-\cos ^2 B\right)\end{aligned}$

$\begin{aligned} & =k^2 \sin ^2 A\left[\left(1-\sin ^2 B\right)-\left(1-\sin ^2 C\right)\right]+k^2 \sin ^2 B\left[\left(1-\sin ^2 C\right)-\left(1-\sin ^2 A\right)\right]+k^2 \sin ^2 C\left[\left(1-\sin ^2 A\right)\right. \\ & \left.-\left(1-\sin ^2 B\right)\right]\end{aligned}$

$\begin{aligned} & =k^2 \sin ^2 A\left(\sin ^2 C-\sin ^2 B\right)+k^2 \sin ^2 B\left(\sin ^2 A-\sin ^2 C\right)+k^2 \sin ^2 C\left(\sin ^2 B-\sin ^2 A\right) \\ & =k^2\left(\sin ^2 A \sin ^2 C-\sin ^2 A \sin ^2 B+\sin ^2 A \sin ^2 B-\sin ^2 B \sin ^2 C+\sin ^2 B \sin ^2 C-\sin ^2 A \sin ^2 C\right) \\ & =k^2(0)=0=\text { RHS. }\end{aligned}$

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=k$

$\therefore a=k \sin A, b=k \sin B$

$\therefore$ (1) gives, $\frac{\cos A}{k \sin A}=\frac{\cos B}{k \sin B}$

$\therefore \frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}$

∴ sin A cos B = cos A sinB ∴ sinA cosB – cosA sinB = 0 ∴ sin (A – B) = 0 = sin0 ∴ A – B = 0 ∴ A = B ∴ the triangle is an isosceles triangle.

$\begin{aligned} & \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=k \\ & \therefore a=k \sin \mathrm{A}, b=k \sin \mathrm{B}, c=k \sin \mathrm{C}\end{aligned}$

$\mathrm{LHS}=\frac{b-c}{a}=\frac{k \sin \mathrm{B}-k \sin \mathrm{C}}{k \sin \mathrm{A}}$

$=\frac{\sin B-\sin C}{\sin A}$

$=\frac{\sin B-\sin C}{\sin \{\pi-(B+C)\}} \quad \ldots[\because A+B+C=\pi]$

$=\frac{\sin B-\sin C}{\sin (B+C)}$

$\begin{aligned} & =\frac{2 \cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{2 \sin \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B+C}{2}\right)} \\ & =\frac{\sin \left(\frac{B-C}{2}\right)}{\sin \left(\frac{B+C}{2}\right)}=\frac{\sin \left(\frac{B}{2}-\frac{C}{2}\right)}{\sin \left(\frac{B}{2}+\frac{C}{2}\right)}\end{aligned}$

$=\frac{\sin \frac{B}{2} \cos \frac{C}{2}-\cos \frac{B}{2} \sin \frac{C}{2}}{\sin \frac{B}{2} \cos \frac{C}{2}+\cos \frac{B}{2} \sin \frac{C}{2}}$

$\sin \frac{B}{2} \cos \frac{C}{2} \quad \cos \frac{B}{2} \sin \frac{C}{2}$

$\cos \frac{B}{2} \cos \frac{C}{2} \quad \cos \frac{B}{2} \cos \frac{C}{2}$

$\frac{\sin \frac{\mathrm{B}}{2} \cos \frac{\mathrm{C}}{2}}{\cos \frac{\mathrm{B}}{2} \cos \frac{\mathrm{C}}{2}}+\frac{\cos \frac{\mathrm{C}}{2} \sin \frac{\mathrm{C}}{2}}{\cos \frac{\mathrm{B}}{2} \cos \frac{\mathrm{C}}{2}}$

$\sin \frac{B}{2} \quad \sin \frac{C}{2}$

$\frac{\sin \frac{B}{2}}{\cos \frac{C}{2}}+\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}$

$=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}=$ RHS.

Alternative Method:

$R H S=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}$

$=\frac{\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}-\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}}{\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}+\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}}$

$\begin{aligned} & =\frac{(s-c) \sqrt{(s-a)}-(s-b) \sqrt{(s-a)}}{(s-c) \sqrt{(s-a)}+(s-b) \sqrt{(s-a)}} \\ & =\frac{(s-c)-(s-b)}{(s-c)+(s-b)}=\frac{s-c-s+b}{s-c+s-b}\end{aligned}$

$\begin{aligned} & =\frac{b-c}{2 s-(b+c)}=\frac{b-c}{(a+b+c)-(b+c)} \\ & =\frac{b-c}{a}=\text { LHS. }\end{aligned}$

$\begin{aligned} & \frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b} \\ & \therefore \frac{\sin ^2 \mathrm{~A}}{a^2}=\frac{\sin ^2 \mathrm{~B}}{b^2}\end{aligned}$

$\mathrm{LHS}=\frac{\cos 2 \mathrm{~A}}{a^2}-\frac{\cos 2 \mathrm{~B}}{b^2}$

$\begin{aligned} & =\frac{1-2 \sin ^2 \mathrm{~A}}{a^2}-\frac{1-2 \sin ^2 \mathrm{~B}}{b^2} \\ & =\frac{1}{a^2}-\frac{2 \sin ^2 \mathrm{~A}}{a^2}-\frac{1}{b^2}+\frac{2 \sin ^2 \mathrm{~B}}{b^2} \\ & =\frac{1}{a^2}-\frac{1}{b^2}-2\left(\frac{\sin ^2 \mathrm{~A}}{a^2}-\frac{\sin ^2 \mathrm{~B}}{b^2}\right)\end{aligned}$

$=\frac{1}{a^2}-\frac{1}{b^2}-2\left(\frac{\sin ^2 \mathrm{~B}}{b^2}-\frac{\sin ^2 \mathrm{~B}}{b^2}\right) \quad \ldots[$ By (1)]

$\begin{aligned} & =\frac{1}{a^2}-\frac{1}{b^2}-2 \times 0 \\ & =\frac{1}{a^2}-\frac{1}{b^2}=\text { RHS }\end{aligned}$

$=\frac{\left(\frac{b^2+c^2-a^2}{2 b c}\right)}{a}+\frac{\left(\frac{c^2+a^2-b^2}{2 a a}\right)}{b}+\frac{\left(\frac{a^2+b^2-c^2}{2 a b}\right)}{c}$

$=\frac{b^2+c^2-a^2}{2 a b c}+\frac{c^2+a^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b c}$

$=\frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2 a b c}$

$=\frac{a^2+b^2+c^2}{2 a b c}=$ RHS

$\begin{aligned} & =a c\left(\frac{c^2+a^2-b^2}{2 c a}\right)-b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =\frac{1}{2}\left(c^2+a^2-b^2\right)-\frac{1}{2}\left(b^2+c^2-a^2\right)\end{aligned}$

$\begin{aligned} & =\frac{1}{2}\left(c^2+a^2-b^2-b^2-c^2+a^2\right) \\ & =\frac{1}{2}\left(2 a^2-2 b^2\right)=a^2-b 2=\text { RHS. }\end{aligned}$

$\begin{aligned} & \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin C}=k \\ & \therefore a=k \sin A, b=k \sin B, c=k \sin C\end{aligned}$

$\begin{aligned} & \text { RHS }=\left(b^2-c^2\right) \sin \mathrm{A} \\ & =\left(k^2 \sin ^2 \mathrm{~B}-k^2 \sin ^2 C\right) \sin \mathrm{A}\end{aligned}$

$\begin{aligned} & =k^2\left(\sin ^2 B-\sin ^2 C\right) \sin A \\ & =k^2(\sin B+\sin C)(\sin B-\sin C) \sin A\end{aligned}$

$\begin{aligned} & =k^2 \times 2 \sin \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right) \times \\ & 2 \cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right) \times \sin A\end{aligned}$

$\begin{aligned} & =k^2 \times 2 \sin \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B+C}{2}\right) \times \\ & 2 \sin \left(\frac{B-C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right) \times \sin A\end{aligned}$

$\begin{aligned} & =k^2 \times \sin (B+C) \times \sin (B-C) \times \sin A \\ & =k^2 \cdot \sin (\pi-A) \cdot \sin (B-C) \cdot \sin A \ldots[\because A+B+C=\pi]\end{aligned}$

$\begin{aligned} & =k^2 \sin A \cdot \sin (B-C) \cdot \sin A \\ & =(k \sin A)^2 \cdot \sin (B-C) \\ & =a^2 \sin (B-C)=L H S .\end{aligned}$

$\begin{aligned} & =\frac{c-\left(\frac{b^2+c^2-a^2}{2 c}\right)}{b-\left(\frac{b^2+c^2-a^2}{2 b}\right)}=\frac{\left(\frac{2 c^2-b^2-c^2+a^2}{2 c}\right)}{\left(\frac{2 b^2-b^2-c^2+a^2}{2 b}\right)} \\ & =\frac{\left(\frac{c^2+a^2-b^2}{2 c}\right)}{\left(\frac{a^2+b^2-c^2}{2 b}\right)}=\frac{\left(\frac{c^2+a^2-b^2}{2 a a}\right)}{\left(\frac{a^2+b^2-c^2}{2 a b}\right)}\end{aligned}$

$=\frac{\cos B}{\cos C}=$ RHS.

$\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ & \therefore a=k \sin A, b=k \sin B, c=k \sin C_c\end{aligned}$

LHS $=a \sin A-b \sin B$

$\begin{aligned} & =k \sin A \cdot \sin A-k \sin B \cdot \sin B \\ & =k\left(\sin ^2 A-\sin ^2 B\right)\end{aligned}$

$=k(\sin A+\sin B)(\sin A-\sin B)$

$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \times \\ & 2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$

$\begin{aligned}=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \times \\ 2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$

$\begin{aligned} & =k \times \sin (A+B) \times \sin (A-B) \\ & =k \sin (\pi-C) \cdot \sin (A-B) \ldots[\because A+B+C=\pi] \\ & =k \sin C \cdot \sin (A-B) \\ & =c \sin (A-B)=\text { RHS. }\end{aligned}$

$\begin{aligned} & =\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2 b^2 c^2}} \\ & =\frac{(s-a)(s-b)(s-c)}{a b c}\end{aligned}$

$\begin{aligned} & =\frac{s(s-a)(s-b)(s-c)}{a b c s} \\ & =\frac{[\mathrm{A}(\triangle \mathrm{ABC})]^2}{a b c s} \ldots[\because \mathrm{A}(\triangle \mathrm{ABC})=\sqrt{s(s-a)(s-b)(s-c)]} \\ & =\text { RHS. }\end{aligned}$

$=(2 s-2 a) \cdot \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

$=2 \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$... (1)

$(c+a-b) \tan \frac{\mathrm{B}}{2}=(a+b+c-2 b) \cdot \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$

$=(2 s-2 b) \cdot \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$

$=2 \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$\ldots(2)$

$(a+b-c) \tan \frac{C}{2}=(a+b+c-2 c) \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$=(2 s-2 c) \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$=2 \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$\ldots(3)$

From (1), (2) and (3), we get

$(b+c-a) \tan \frac{\mathrm{A}}{2}=(c+a-b) \tan \frac{\mathrm{B}}{2}=(a+b-c) \tan \frac{\mathrm{C}}{2}$

$=a\left(2 \sin ^2 \frac{C}{2}\right)+c\left(2 \sin ^2 \frac{\mathrm{A}}{2}\right)$

$=a(1-\cos \mathrm{C})+c(1-\cos \mathrm{A})$

$=a\left[1-\frac{a^2+b^2-c^2}{2 a b}\right]+c\left[1-\frac{b^2+c^2-a^2}{2 b c}\right]$

... [By cosine rule]

$=a\left[\frac{2 a b-a^2-b^2+c^2}{2 a b}\right]+c\left[\frac{2 b c-b^2-c^2+a^2}{2 b c}\right]$

$=\frac{2 a b-a^2-b^2+c^2}{2 b}+\frac{2 b c-b^2-c^2+a^2}{2 b}$

$=\frac{2 a b-a^2-b^2+c^2+2 b c-b^2-c^2+a^2}{2 b}$

$=\frac{2 a b-2 b^2+2 b c}{2 b}$

$=a-b+c=$ RHS

$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=k$

$\therefore a=k \sin \mathrm{A}, b=k \sin \mathrm{B}, c=k \sin \mathrm{C}$

$\mathrm{RHS}=\left(\frac{b-c}{a}\right) \cos \frac{\mathrm{A}}{2}$

$=\left(\frac{k \sin \mathrm{B}-k \sin \mathrm{C}}{k \sin \mathrm{A}}\right) \cos \frac{\mathrm{A}}{2}$$=\left(\frac{k \sin \mathrm{B}-k \sin \mathrm{C}}{k \sin \mathrm{A}}\right) \cos \frac{\mathrm{A}}{2}$

$=\left(\frac{\sin B-\sin C}{\sin A}\right) \cos \frac{A}{2}$

$=\frac{2 \cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}} \cdot \cos \frac{A}{2}$

$=\frac{\cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$

$=\frac{\cos \left(\frac{\pi}{2}-\frac{A}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}} \ldots[\because A+B+C=\pi]$

$=\frac{\sin \frac{A}{2} \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$

$=\sin \left(\frac{B-C}{2}\right)=$ LHS

$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin C}$

$\therefore \frac{a}{b}=\frac{\sin A}{\sin B}$ and $\frac{b}{c}=\frac{\sin B}{\sin C}$

$\therefore a: b: c=\sin A: \sin B: \sin C$

Given $\angle A=45^{\circ}$ and $\angle B=60^{\circ}$

$\because \angle A+\angle B+\angle C=180^{\circ}$

$\therefore 45^{\circ}+60^{\circ}+\angle C=180^{\circ}$

$\therefore \angle C=180^{\circ}-105^{\circ}=75^{\circ}$

Now, $\sin A=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

$\sin B=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

and $\sin C=\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$

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

$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin C}$

$\therefore \frac{a}{b}=\frac{\sin A}{\sin B}$ and $\frac{b}{c}=\frac{\sin B}{\sin C}$

$\therefore a: b: c=\sin A: \sin B: \sin C$

Given $\angle A=45^{\circ}$ and $\angle B=60^{\circ}$

$\because \angle A+\angle B+\angle C=180^{\circ}$

$\therefore 45^{\circ}+60^{\circ}+\angle C=180^{\circ}$

$\therefore \angle C=180^{\circ}-105^{\circ}=75^{\circ}$

Now, $\sin A=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

$\sin B=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

and $\sin C=\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$

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

$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

$\therefore$ the ratio of the sides of $\triangle A B C$

$=a: b: c=\sin \mathrm{A}: \sin \mathrm{B}: \sin \mathrm{C}$

$\therefore a: b: c=2: \sqrt{6}:(\sqrt{3}+1)$

$=\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+1}{2 \sqrt{2}}$

$\begin{aligned} & \tan \frac{\pi}{4}=1 \text { and } \tan (\pi-\theta)=-\tan \theta \\ & \tan (2 \pi-\theta)=-\tan \theta\end{aligned}$

$\begin{aligned} & \therefore \tan \left(\pi-\frac{\pi}{4}\right)=-\tan \frac{\pi}{4}=-1 \\ & \text { and } \tan \left(2 \pi-\frac{\pi}{4}\right)=-\tan \frac{\pi}{4}=-1\end{aligned}$

$\begin{aligned} & \therefore \tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}=-1, \text { where } \\ & 0<\frac{3 \pi}{4}<2 \pi \text { and } 0<\frac{7 \pi}{4}<2 \pi\end{aligned}$

$\begin{aligned} & \therefore \tan \theta=-1 \text { gives, } \\ & \tan \theta=\tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}\end{aligned}$

$\therefore \theta=\frac{3 \pi}{4}$ and $\theta=\frac{7 \pi}{4}$

Hence, the required principal solutions are

$\theta=\frac{3 \pi}{4}$ and $\theta=\frac{7 \pi}{4}$.

$\begin{aligned} & \sin \frac{\pi}{6}=\frac{1}{2} \text { and } \sin (\pi+\theta)=-\sin \theta \\ & \sin (2 \pi-\theta)=-\sin \theta\end{aligned}$

$\begin{aligned} & \therefore \sin \left(\pi+\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}=-\frac{1}{2} \\ & \text { and } \sin \left(2 \pi-\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}=-\frac{1}{2}\end{aligned}$

$\therefore \sin \frac{7 \pi}{6}=\sin \frac{11 \pi}{6}=-\frac{1}{2}$, where

$0<\frac{7 \pi}{6}<2 \pi$ and $0<\frac{11 \pi}{6}<2 \pi$

$\therefore \sin \theta=-\frac{1}{2}$ gives,

$\sin \theta=\sin \frac{7 \pi}{6}=\sin \frac{11 \pi}{6}$

$\therefore \theta=\frac{7 \pi}{6}$ and $\theta=\frac{11 \pi}{6}$

Hence, the required principal solutions are

$\theta=\frac{7 \pi}{6}$ and $\theta=\frac{11 \pi}{6}$

$\theta=2 n \pi \pm \alpha, n \in Z$

∴ the general solution of cos 4θ = cos 2θ is given by

$4 \theta=2 n \pi \pm 2 \theta, n \in Z$

Taking positive sign, we get

$\begin{aligned} & 4 \theta=2 n \pi+2 \theta, n \in Z \\ & \therefore 2 \theta=2 n \pi, n \in Z\end{aligned}$

$\therefore \theta=n \pi, n \in Z$

Taking negative sign, we get

$\begin{aligned} & 4 \theta=2 n \pi-2 \theta, n \in Z \\ & \therefore 6 \theta=2 n \pi, n \in Z \\ & \therefore \theta=\frac{n \pi}{3}, n \in Z\end{aligned}$

Hence, the required general solution is

$\theta=\frac{n \pi}{3}, n \in Z$ or $\therefore \theta=n \pi, n \in Z$.

$\begin{aligned} & \cos 4 \theta=\cos 2 \theta \\ & \therefore \cos 4 \theta-\cos 20=0 \\ & \therefore-2 \sin \left(\frac{4 \theta+2 \theta}{2}\right) \cdot \sin \left(\frac{4 \theta-2 \theta}{2}\right)=0\end{aligned}$

$\begin{aligned} & \therefore \sin 3 \theta \cdot \sin \theta=0 \\ & \therefore \text { either } \sin 3 \theta=0 \text { or } \sin \theta=0\end{aligned}$

The general solution of sin θ = 0 is

$\theta=n \pi, n \in Z$.

∴ the required general solution is given by

$\begin{aligned} & 3 \theta=n \pi, n \in Z \text { or } \theta=n \pi, n \in Z \\ & \text { i.e. } \theta=\frac{n \pi}{3}, n \in Z \text { or } \theta=n \pi, n \in Z\end{aligned}$

$\theta=n \pi \pm \alpha, n \in Z$

Now, $4 \sin ^2 \theta=3$

$\therefore \sin ^2 \theta=\frac{1}{4}=\left(\frac{1}{2}\right)^2$

$\therefore \sin ^2 \theta=\left(\sin \frac{\pi}{6}\right)^2 \quad \cdots\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]$

$\therefore \sin ^2 \theta=\sin ^2 \frac{\pi}{6}$

$\therefore$ the required general solution is $\theta=n \pi \pm \frac{\pi}{6}, n \in Z$.

$\theta=n \pi \pm \alpha, n \in Z$ Now, $4 \cos ^2 \theta=3$

$\therefore \cos ^2 \theta=\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2$

$\therefore \cos ^2 \theta=\left(\cos \frac{\pi}{6}\right)^2 \quad \cdots\left[\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]$

$\therefore \cos ^2 \theta=\cos ^2 \frac{\pi}{6}$

$\therefore$ the required general solution is given by

$\theta=n \pi \pm \frac{\pi}{6}, n \in Z$

$\theta=n \pi+\alpha, n \in Z$

Now, $\cot 4 \theta=-1$

$\therefore \tan 4 \theta=-1$

$\therefore \tan 4 \theta=-\tan \frac{\pi}{4} \quad \cdots\left[\because \tan \frac{\pi}{4}=1\right]$

$\therefore \tan 4 \theta=\tan \left(\pi-\frac{\pi}{4}\right) \ldots[\because \tan (\pi-\theta)=-\tan \theta]$

$\therefore \tan 4 \theta=\tan \frac{3 \pi}{4}$

$\therefore$ the required general solution is given by

$4 \theta=n \pi+\frac{3 \pi}{4}, n \in Z$

i.e. $\theta=\frac{n \pi}{4}+\frac{3 \pi}{16}, n \in Z$.

$\theta=n \pi+\alpha, n \in Z$

Now, $\tan \frac{2 \theta}{3}=\sqrt{3}$

$\therefore \tan \frac{2 \theta}{3}=\tan \frac{\pi}{3} \quad \cdots\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]$

$\therefore$ the required general solution is given by

$\frac{2 \theta}{3}=n \pi+\frac{\pi}{3}, n \in Z$

i.e. $\theta=\frac{3 n \pi}{2}+\frac{\pi}{2}, n \in Z$.

$\theta=n \pi+(-1)^n \alpha, n \in Z$

Now, $\sin 2 \theta=\frac{1}{2}$

$\therefore \sin 2 \theta=\sin \frac{\pi}{6} \quad \cdots\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]$

$\therefore$ the required general solution is given by

$2 \theta=n \pi+(-1)^n\left(\frac{\pi}{6}\right), n \in Z$

i.e. $\theta=\frac{n \pi}{2}+(-1)^n\left(\frac{\pi}{12}\right), n \in Z$.