LHS\(=\cos^{-1}\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)\)
put \(\frac{3}{5}=\cos\theta\)
\(\therefore\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\left(\frac35\right)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac45\)
Now, \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac45}{\frac35}=\frac43\Rightarrow\theta=\tan^{-1}\left(\frac43\right)\)
\(\therefore\cos^{-1}\left(\frac{3}{5}\cos x+\frac45\sin x\right)\)
= cos^-1 (cos \(\theta\) cos x + sin \(\theta\) sin x)
= cos^-1 [cos (x - \(\theta\) )]    [\(\therefore\) cos (A- B) = cos A cos B + sin A sin B]
= x - \(\theta\)
= \(x-{\tan}^{-1}\left({{4}\over{3}} \right)\) = RHS
Hence proved.

LHS\(=\frac{\sin5x-2\sin3x+\sin x}{\cos 5x-\cos x}\)
\(=\frac{(\sin5x+\sin x)-2\sin3x}{\cos5x-\cos x}\)
\(=\frac{2\sin\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right)-2\sin3x}{-2\sin\left(\frac{5x+x}{2}\right)\sin\left(\frac{5x-x}{2}\right)}\)
\(=\frac{2\sin3x\cos2x-2\sin3x}{-2\sin3x.\sin2x}\)
\(\frac{-2\sin3x(1-\cos2x)}{-2\sin3x.\sin2x}=\frac{1-\cos2x}{\sin2x}\)
\(=\frac{2\sin^2x}{2\sin x\cos x}=\frac{\sin x}{\cos x}\tan x=RHS\)
Hence proved.

LHS\(=\frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}\)
\(\left[\because\sin C\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\ and\ \cos C\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\right]\)
\(=\frac{2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)}{{2\cos\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)}}=\frac{2\sin4x.\cos x}{2\cos4x.\cos x}=\tan4x=RHS\)
Hence proved.

LHS\(=\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)\)
\(=\frac{2\sin\left(\frac{\pi}{3}+x\right).\sin\left(\frac{\pi}{3}-x\right)}{2\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)}\)
\([\because2\sin A\sin B=\cos(A-B)-\cos(A+B)\ and\ \ 2\cos A\cos B=\cos(A+B)+\cos(A-B)]\)
\(=\frac{\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)}{\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)+\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)}\)
\(=\frac{\cos2x-\cos\frac{2\pi}{3}}{\cos\frac{2\pi}{3}+\cos2x}=\frac{\cos2x+\frac{1}{2}}{-\frac12+\cos2x}\)
\(=\frac{2\cos2x+1}{2\cos2x-1}\)
=RHS
Hence proved.

Consider \(\tan(\alpha+\beta)\)
\(=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\left(\frac{m}{m+1}\right)\left(\frac{1}{2m+1}\right)}=\frac{\frac{m(2m+1)+1(m+1)}{(m+1)(2m+1)}}{\frac{(m+1)(2m+1)-m}{(m+1)(2m+1)}}\)
\(=\frac{\frac{2m^2+m+m1}{(m+1)(2m+1)}}{\frac{2m^2+m+2m+1-m}{(m+1)(2m+1)}}=\frac{2m^2+2m+1}{(m+1)(2m+1)}\times\frac{(m+1)(2m+1)}{2m^2+2m+1}\)
\(=1=\tan\frac{\pi}{4}\therefore\tan(\alpha+\beta)=\tan\frac{\pi}{4}\Rightarrow\alpha+\beta=\frac{\pi}{4}\)

LHS=\(\frac{\sin(x+y)}{\sin(x-y)}=\frac{\sin x\cos y+\cos x+sin y}{\sin x\cos y-\cos x\sin y}\)
Dividing the numerator and denominator by cos x cos y,
We get LHS,
\(\frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y}{\cos x\cos y}-\frac{\cos x\sin y}{\cos x\cos y}}=\frac{\tan x+\tan y}{\tan x-\tan y}=RHS\)
Hence proved.

Given \(\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) }\)
put \(​​x=\sec { \theta \Rightarrow } \theta =\sec ^{ -1 }{ \left( x \right) } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { \sec ^{ 2 }{ \theta -1 } } } \right) } =\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { \tan ^{ 2 }{ \theta } } } \right) } \)   \([\because sec^{ 2 }\theta -1=tan^{ 2 }\theta ]\)
\(=\tan ^{ -1 }{ \left( \frac { 1 }{ \tan { \theta } } \right) } =\tan ^{ -1 }{ \left( \cot { \theta } \right) } =\tan ^{ -1 }{ \left( \frac { \pi }{ 2 } -\theta \right) } \)
\(=\frac { \pi }{ 2 } -\theta =\frac { \pi }{ 2 } -\sec ^{ -1 }{ \left( x \right) } \)
\(\therefore \tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) } =\frac { \pi }{ 2 } -\sec ^{ -1 }{ \left( x \right) } \) which is the simplest form.

LHS\(=\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}\)
=(1 + cot x + cosec x) (1 + cot x - cosec x)
(1 + cot x)² - (cosec²x) [\(\therefore\)(a + b) (a - b) = a² - b²]
1 + cot²x + 2 cot x - cosec²x
= cosec²x + 2 cot x - cosec²x [\(\therefore\) 1 + cot²x = cosec²x]
= 2 cot x = RHS. Hence proved.

LHS\(=\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}\)
\(=\left(sin\frac{\pi}{6}\right)^2+\left(\cos\frac{\pi}{3}\right)^2-\left(\tan\frac{\pi}{4}\right)^2=\left(\frac12\right)^2+\left(\frac{1}{2}\right)^2-1^2=\frac{1}{4}+\frac{1}{4}-1\)
\(=\frac{1+1-4}{4}=-\frac24=-\frac12\)=RHS
Hence proved.

We know that \(\cos^2x+\sin^2x=1\)
\(\Rightarrow\cos x=\pm\sqrt{1-\sin^2x}\)
In the III quadrant, cos x is negative
\(\therefore\cos x=-\sqrt{1-\sin^2x}=-\sqrt{1-\frac{24}{25}}=-\frac15\left[\because\sin^2x=\frac{4(6)}{25}=\frac{24}{25}\right]\)
In the III quadrant, tan x is positive
\(\therefore\tan x=\frac{\sin x}{\cos x}=\frac{-2\sqrt6}{5}\times\frac{-5}{1}=2\sqrt{6}\)
\(cosec x=\frac{1}{\sin x}=\frac{-5}{2\sqrt6}\)
\(\sec x=\frac{1}{\cos x}=-5\) and
\(\cot x=\frac{1}{\tan x}=\frac{1}{2\sqrt6}\)