Maths Solve the Following Question.(4 Marks)

Given, $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\cot ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$;
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)=\frac{\pi}{4}-\tan ^{-1}\left(\frac{x+1}{x+2}\right)$
$=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{x+1}{x+2}\right)$
$=\tan ^{-1}\left[\frac{1-\frac{x+1}{x+2}}{1+\frac{x+1}{x+2}}\right] \ldots \ldots \ldots .\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$=\tan ^{-1}\left[\frac{x+2-x-1}{x+2+x+1}\right]$
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)=\tan ^{-1}\left(\frac{1}{2 x+3}\right)$
$\frac{x-1}{x-2}=\frac{1}{2 x+3}$
$(x-1)(2 x+3)=x-2$
$2 x^2-1=0$
$2 x^2=1$
$x^2=\frac{1}{2}$
$x= \pm \frac{1}{\sqrt{2}}$

Let ‘a’ be the edge of the cube with one vertex at the origin.
The diagonals of the cube are OG, AD, CF and EB.
Consider the diagonals OG and AD.
The d.r.s of OG are
a - 0, a - 0, a - 0 = a, a, a
The d.r.s of AD are
0 - a, a -0, a -0 = a, a, a [1]
Let θ be the angle between OG and AD
$\cos \theta=\frac{a(-a)+a(a)+a(a)}{\sqrt{a^2+a^2+a^2} \cdot \sqrt{(-a)^2+a^2+a^2}}$
$=\frac{a^2}{3 a^2}$
$\cos \theta=\frac{1}{3}$
$\theta=\cos ^{-1}\left(\frac{1}{3}\right)$

In ΔABC by sine rule, we have
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}= k$
∴ a = k sin A, b = k sin B and c = k sin C
Now, consider
$\frac{a-b}{a+b}=\frac{k \sin A-k \sin B}{k \sin A+k \sin B}$
$=\frac{\sin A-\sin B}{\sin A+\sin B}$
$=\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$
$=\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)$
$=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right) \ldots \ldots[\because A + B + C =\pi]$
$=\tan \left(\frac{C}{2}\right) \tan \left(\frac{A-B}{2}\right)$
$\therefore \frac{a-b}{a+b}=\tan \frac{C}{2} \cdot \tan \left(\frac{A-B}{2}\right)$
$\therefore \tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b} \cot \left(\frac{C}{2}\right)$

sinx · tanx = tanx - sinx + 1
$\sin x \cdot\left(\frac{\sin x}{\cos x}\right)-1=\frac{\sin x}{\cos x}-\sin x$
$\frac{\sin ^2 x-\cos x}{\cos x}=\frac{\sin x-\sin x \cdot \cos x}{\cos x}$
$\sin ^2 x-\cos x=\sin x-\sin x \cdot \cos x$
now $x$ can't equal to $\frac{\pi}{2}$
$\sin ^2 x-\cos x-\sin x+\sin x \cos x=0$
$\sin ^2 x+\sin x \cos x-\cos x-\sin x=0$
$\sin x(\sin x+\cos x)-1(\sin x+\cos x)=0$
$(\sin x-1)(\sin x+\cos x)=0$
since $x$ is not $\frac{\pi}{2}$
$\sin x+\cos x=0$
$\sin x=-\cos x$
$x=$ multiple of $\frac{\pi}{4}$ in even quads.

Let $\cos ^{-1} \frac{12}{13}=x$
$\therefore \cos x=\frac{12}{13}$
$\therefore \sin x=\frac{5}{13}$
and let $\sin ^{-1} \frac{3}{5}=y$
$\sin y=\frac{3}{5}$
$\therefore \cos y=\frac{4}{5}$
∴ using sin (x + y) = sin x cos y + cos x sin y
$=\frac{5}{13} \times \frac{4}{5}+\frac{12}{13} \times \frac{3}{5}$
$=\frac{20+36}{13 \times 5}$
$=\frac{56}{65}$
$\therefore x+y=\sin ^{-1} \frac{56}{65}$
$\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$
Hence proved.

$LHS =(a-b)^2 \cos ^2\left(\frac{C}{2}\right)+(a+b)^2 \sin ^2\left(\frac{C}{2}\right)$
$=a^2\left[\cos ^2\left(\frac{C}{2}\right)+\sin ^2\left(\frac{C}{2}\right)\right]+b^2\left[\cos ^2\left(\frac{C}{2}\right)+\sin ^2\left(\frac{C}{2}\right)\right]-2 a b\left[\cos ^2\left(\frac{C}{2}\right)-\sin ^2\left(\frac{C}{2}\right)\right]$
$=a^2+b^2-a^2-b^2+c^2$
$=c^2$
=RHS
Hence proved

cos x - sin x = 1
Dividing by $\sqrt{1^2+(-1)}=\sqrt{2}$
$\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x=\frac{1}{\sqrt{2}}$
$\cos \frac{\pi}{4} \cos x -\sin \frac{\pi}{4} \sin x =\frac{1}{\sqrt{2}}$
$\cos \left( x +\frac{\pi}{4}\right)=\cos \frac{\pi}{4} \quad \ldots$ (i)
The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \frac{\pi}{4} ; n \in z$
∴ The genera; solution of equation (i) given by 
$x +\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4} ; n \in z$
$x =2 n \pi ; x=2 n \pi-\frac{\pi}{2} ; n \in z$

$(\sin 2 x+\sin 6 x)+\sin 4 x=0$
$2 \sin 4 x \cdot \cos 2 x+\sin 4 x=0$
$\sin 4 x(2 \cos 2 x+1)=0$
$\sin 4 x=0$ or $2 \cos 2 x+1=0$
$\sin 4 x=0$ or $\cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)$
Using $\sin x=0 \Rightarrow x=n \pi$
$\sin 4 x=0$
$4 x=n \pi$
The genral solution is $x$
$x=\frac{n \pi}{4}$
using $\cos x=\cos \alpha \Rightarrow x=2 m x \pm \alpha$
$\cos 2 x=\cos \left(\frac{2 \pi}{3}\right)$
$2 x=2 \mp i \pm \frac{2 \pi}{3}$
The genral solution is x
$x=\mp i \pm \frac{\pi}{3}$ where $m, n \in z$

$\angle A, \angle B, \angle C$ are in A.P and $b: c=\sqrt{3}: \sqrt{2}$
$\therefore 2 B=A+C$
$2 B=180^{\circ}-B \quad \ldots \ldots \ldots \ldots \ldots . .\left[\because A+B+C=180^{\circ}\right]$
$3 B=180^{\circ}$
$\angle B=60^{\circ}$
In ΔABC by sine rule, we have
$\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{\sin B}{\sin C}=\frac{b}{c}$
$\frac{\sin 60^{\circ}}{\sin c}=\frac{\sqrt{3}}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2}}=\sin C$
$\sin C=\frac{1}{\sqrt{2}}$
$\angle C=45^{\circ}$
$\angle A=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$
Thus, the angles of $\triangle ABC$ are $\angle A=75^{\circ}, \angle B=60^{\circ}, \angle C=45^{\circ}$