Firstly, expand the terms in left-hand side of the equation and simplify them. Then, compare the value of LHS with RHS of identity to get required values.
p = 3,  q = - 3

LHS\(=\sec ^{ 2 }{ \theta } -\frac { \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 4 }{ \theta } } -\frac { 2\sin ^{ 4 }{ \theta } }{ \cos ^{ 4 }{ \theta } } }{ \frac { 2\cos ^{ 4 }{ \theta } }{ \cos ^{ 4 }{ \theta } } -\frac { \cos ^{ 2 }{ \theta } }{ \cos ^{ 4 }{ \theta } } } \)
\(=\sec ^{ 2 }{ \theta } -\frac { \tan ^{ 2 }{ \theta } .\sec ^{ 2 }{ \theta } -2\tan ^{ 4 }{ \theta } }{ 2-\sec ^{ 2 }{ \theta } } \)
\(=\sec ^{ 2 }{ \theta } -\frac { \tan ^{ 2 }{ \theta } (\sec ^{ 2 }{ \theta } -2\tan ^{ 2 }{ \theta } ) }{ 2-\sec ^{ 2 }{ \theta } } \)
\(=\sec ^{ 2 }{ \theta } -\frac { \tan ^{ 2 }{ \theta } (2-\sec ^{ 2 }{ \theta } ) }{ 2-\sec ^{ 2 }{ \theta } } =\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } =1\)

\(\cot ^{ 2 }{ \theta } \tan ^{ 2 }{ \theta } =\frac { 1 }{ \tan ^{ 2 }{ \theta } } .\tan ^{ 2 }{ \theta } =1\)

\(A+B+C={ 180 }^{ 0 }\Rightarrow \frac { B+C }{ 2 } ={ 90 }^{ 0 }-\frac { A }{ 2 } \Rightarrow \tan { \left( \frac { B+C }{ 2 } \right) } =\tan { \left( { 90 }^{ 0 }-\frac { A }{ 2 } \right) } \Rightarrow \tan { \frac { B+C }{ 2 } } =\cot { \frac { A }{ 2 } } \)

x = 3

\(\sin { \theta } -\cos { \theta } =0\Rightarrow \sin { \theta } =\cos { \theta } \Rightarrow \tan { \theta } =1\Rightarrow \theta ={ 45 }^{ 0 }\)

(i) \(\frac { \sqrt { 3 } }{ 4 } \)
(ii) 4

As, cosA = \(\frac {8}{17} =\frac {B}{H}\)
Using Pythagoras theorem, P² + 8² = 17² \(\Rightarrow\) P = 15.
Then, 15 cosec A - 8 sec B
= 15 \( \frac {H}{P} \)- 8 \( \frac {H}{B}\) = 15 \( (\frac {17}{15})\)- 8 \( (\frac {17}{8} )\) = 17 - 17=  0

\(\frac { \sqrt { 5 } }{ 3 } \)

Use Pythagoras theorem, to find base, then on the basis of angle find sin A.
\(\frac {8}{17}\)