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
\(\cot x=\frac{1}{\tan x}=\frac{1}{2\sqrt6}\)