Maths Solve the Following Question.(2 Marks)

Consider L.H.S. $=\frac{a-b \cos C}{b-a \cos C}$
$=\frac{b \cos C+c \cos B-b \cos C}{a \cos C+c \cos A-a \cos C}$
$=\frac{c \cos B}{c \cos A}$
$=\frac{\cos B}{\cos A}=R . H . S$
$\therefore \frac{a-b \cos C}{b-a \cos C}=\frac{\cos B}{\cos A}$

$\sec x=\frac{2}{\sqrt{3}}$
$\cos x=\frac{\sqrt{3}}{2}=\frac{\cos \pi}{6}=\cos (2 \pi-\pi 6)$
$\frac{\pi}{6}, \frac{11 \pi}{6}$

(B) $\sqrt{\frac{1}{5}}$
$s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{(21-14)(21-15)}{14 \times 15}}=\sqrt{\frac{1}{5}}$

$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
$\therefore \sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x$
$\therefore 1-x=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x \right)$
$\therefore 1- x =\cos \left(2 \sin ^{-1} x \right) \ldots\left[\cdot\left[\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta\right]\right.$
$\therefore 1-x=1-2\left[\sin \left(\sin ^{-1} x\right)\right]^2 \quad \ldots .\left[\because \cos 2 \theta=1-2 \sin ^2 \theta\right]$
$\therefore 1-x=1-2 x^2$
$\therefore 2 x^2-x=0$
$\therefore x(2 x-1)=0$
$\therefore x=0$ or $x=\frac{1}{2}$
When $x=\frac{1}{2}$
LHS $=\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)$
$=\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)$
$=-\sin ^{-1}\left(\frac{1}{2}\right)$
$=-\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$
$=-\frac{\pi}{6} \neq \frac{\pi}{2}$
$\therefore x \neq \frac{1}{2}$
Hence, x = 0.

The principal solution of $\cos ^{-1}\left(-\frac{1}{2}\right)=$ An angle in $[0, \pi]$, whose cosine is $-1 / 2$
$\Rightarrow \cos ^{-1}\left(-\frac{1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right) \quad \ldots \ldots\left[\right.$ [because $\left.\cos ^{-1}(-x)=\pi-\cos x\right]$
$=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

$\tan ^2 \theta=1=\tan ^2\left(\frac{\pi}{4}\right)$
$\tan ^2 \theta=\tan ^2 \alpha \Rightarrow \theta=n \pi \pm \alpha$
$\therefore \theta=n \pi \pm \frac{\pi}{4}$

cos x + sinx = 1
$\cos x \frac{1}{\sqrt{2}}+\sin x \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\cos x \frac{\cos \pi}{4}+\sin x \frac{\sin \pi}{4}=\frac{\cos \pi}{4}$
$\cos \left(x-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right)$
by $\cos y=\cos \alpha$
$y=2 n \pi \pm \alpha$
$x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}$
$x-\frac{\pi}{4}=2 n \pi+\frac{\pi}{4}$ or $x-\frac{\pi}{4}=2 n \pi-\frac{\pi}{4}$
$x=2 n \pi+\frac{\pi}{2}$ or $x=2 n \pi$

Given:
$4 \cos ^2 x=1$
$\cos ^2 x=\frac{1}{4}$
$\cos ^2 x=\cos ^2\left(\frac{\pi}{3}\right)$
$\left[\right.$ Using $\left.\cos ^2 x=\cos ^2 \alpha \Rightarrow x=n \pi \pm \alpha\right]$
$x=n \pi \pm \frac{\pi}{3}$ where $n \in Z$

$\frac{5 \pi}{6}$
$\cot x=-\sqrt{3}$
$\cot x=-\cot x$
we know $\cot x$;is negative in 2nd and 4th quadrant so, we convert this 2nd or 4th
now
$\cot x=-\cot x=\cot \left(\pi-\frac{5 \pi}{6}\right)$
$x=\frac{5 \pi}{6}$
also
$\cot x=\cot \left(2 \pi-\frac{11 \pi}{6}\right)$
$x=\frac{11 \pi}{6}$
only option given is $\frac{5 \pi}{6}$

(B) $\frac{1}{\sqrt{10}}$
$s=\frac{a+b+c}{2}=\frac{18+24+30}{2}=36$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{(36-24)(36-30)}{24 \times 30}}=\sqrt{\frac{12 \times 6}{24 \times 30}}=\frac{1}{\sqrt{10}}$