Multiplying numerator and denominator by 2, we have
LHS\(=\frac{(2sin11AsinA)+(2sin7Asin3A)}{(2cos11AsinA)+(2cos7Asin3A)}\)
\(=\frac{(cos10A-cos12A)+(cos4A-cos10A)}{(sin12A-sin10A)+(sin10A-sin4A)}\)
\(=\frac{cos4A-cos12A}{sin12A-sin4A}\)
\(=\frac{2sin8Asin4A}{2cos8Asin4A}=tan8A=RHS\)

= cos 20° cos 40° \((\frac{1}{2})\)cos 80°
\(=\frac{1}{4}\)(cos 20° cos 40° cos 80°)
\(=\frac{1}2(\frac{1}{8})\)
\(=\frac{1}{16}\)

LHs = 4 cos 12° cos 48° cos 72°
= 2 ( 2 cos 12° cos 48°) cos 72°
= 2 ( cos 60° + cos 36° ) cos 72°
= 2 cos 60° cis 72° + 2 cos 36° cos 72°    [ \(\because\) 2 cos A cos B = cos ( A + B ) + cos ( A - B ) ]
= \(2\times{{1}\over{2}}\) cos 72° + 2 cos 36° cos 72°
= cos 72° + 2 cos 36° cos 72°
= cos 72° + cos (108°) + cos 36°   [ \(\because\) 2 cos A cos B = cos ( A + B ) + cos ( A - B ) ]
= cos 72° + cos (108-72°) + cos 36°
= cos 72°-cos 72°+cos 36°  [ \(\because\) cos (180-\(\theta\)) =-cos \(\theta\)]
= cos 36°
= RHS
Hence, proved.

Let \(x=a\sin { \theta } \Rightarrow \frac { x }{ a } =\sin { \theta } \Rightarrow \sin ^{ -1 }{ \left( \frac { x }{ a } \right) } \)
LHS = \(\tan ^{ -1 }{ \left( \frac { a\sin { \theta } }{ \sqrt { { a }^{ 2 }-{ a }^{ 2 }\sin ^{ 2 }{ \theta } } } \right) } =\tan ^{ -1 }{ \left( \frac { a\sin { \theta } }{ a\sqrt { 1-\sin ^{ 2 }{ \theta } } } \right) } \)
= \(\tan ^{ -1 }{ \left( \frac { \sin { \theta } }{ \sqrt { \cos ^{ 2 }{ \theta } } } \right) } \)
= \(\tan ^{ -1 }{ \left( \frac { \sin { \theta } }{ \cos { \theta } } \right) } \)
= \(\tan ^{ -1 }{ \left( \tan { \theta } \right) } =\theta =\sin ^{ -1 }{ \left( \frac { x }{ a } \right) } \)

Given sin 2x + cos x = 0
cos x = -sin 2x
\(cosx=cos\left( \frac { \pi }{ 2 } +2x \right) \)
\(x=2n\pi \pm \left( \frac { \pi }{ 2 } +2x \right) ,n\in z\)
Taking positive sign we get
\(x=2n\pi +\left( \frac { \pi }{ 2 } +2x \right) \)
\(\Rightarrow -x=2n\pi +\frac { \pi }{ 2 } ,n\in z\)
\(\Rightarrow x=2n\pi -\frac { \pi }{ 2 } \) where m = -n∈z
Taking negative sign we get,
\(x=2n\pi -\left( \frac { \pi }{ 2 } +2x \right) \)
\(\Rightarrow 3x=2n\pi -\frac { \pi }{ 2 } \)
\(x=\frac { 2n\pi }{ 3 } -\frac { \pi }{ 6 } ,n\epsilon z\)
Hence \(x=2m\pi -\frac { \pi }{ 2 } \) or \(\frac { 2n\pi }{ 3 } -\frac { \pi }{ 6 } \) where m, n∈z