LHS = cos²2x - cos²6x
= (cos 2x + cos 6x)(cos 2x - cos 6x) [∴ cos²A - cos²B = (cos A + cos B)(cos A - cos B)
\(=\left[ 2cos\left( \frac { 2x+6x }{ 2 } \right) .cos\left( \frac { 2x-6x }{ 2 } \right) \right] \left[ -2sin\left( \frac { 2x+6x }{ 2 } \right) .sin\left( \frac { 2x-6x }{ 2 } \right) \right] \)
\(=\left[ \because cosC+cosD=2\quad cos\left( \frac { C+D }{ 2 } \right) cos\left( \frac { C-D }{ 2 } \right) and\quad cosC-cosD=-2sin\left( \frac { C+D }{ 2 } \right) .sin\left( \frac { C-D }{ 2 } \right) \right] \)
= [2 cos 4x . cos(-2x)][-2 sin 4x . sin(-2x)]
= (2 cos 4x . cos 2.x) (2 sin4x . sin 2.x) [:. cos (-\(\theta\)) = cos \(\theta\) and sin (-\(\theta\)) = -sin\(\theta\)]
= (2 sin 2.x cos 2x) (2 sin 4x cos 4x)
= sin 4x sin 8x [sin 2A = 2 sin A cos A] = RHS.
Hence proved

Since x lies in the II quadrant ,cos x is negative
∴ cos x = \(-\left( \frac { 1 }{ 1+\sqrt { { tan }^{ 2 }x } } \right) =\frac { -1 }{ \sqrt { 1+\frac { 16 }{ 9 } } } =\frac { -3 }{ 5 } \)
Now \(\frac { \pi }{ 2 }     lies in I quadrant \(sin\frac { x }{ 2 } ,cos\frac { x }{ 2 } \) \(tan\frac { x }{ 2 } \) are positive
\(\therefore cos\frac { x }{ 2 } =\sqrt { \frac { 1+cos\quad x }{ 2 } } =\sqrt { \frac { 1-\frac { 3 }{ 5 } }{ 2 } } =\sqrt { \frac { 5-3 }{ 2\times 5 } } =\sqrt { \frac { 1 }{ 5 } } =\frac { 1 }{ \sqrt { 5 } } \)
\(sin\frac { x }{ 2 } =\sqrt { \frac { 1-cos\quad x }{ 2 } } =\sqrt { \frac { 1+\frac { 3 }{ 5 } }{ 2 } } =\sqrt { \frac { 5+3 }{ 2\times 5 } } =\sqrt { \frac { 4 }{ 5 } } =\frac { 2 }{ \sqrt { 5 } } \)
\(tan\frac { x }{ 2 } =tan\frac { x }{ 2 } =\sqrt { \frac { 1-cos\quad x }{ 1+cos\quad x } } =\sqrt { \frac { 1+\frac { 3 }{ 5 } }{ 1-\frac { 3 }{ 5 } } } =\sqrt { \frac { 8 }{ 2 } } =\sqrt { 4 } =2\)

LHS = sin(n + 1)x sin(n + 2) x + cos(n + 1)xcos(n + 2)x
Let A = (n + 1)x and B = (n + 2)x
= sinAsinB + cosAcosB = cos(A - B)
= cos[(n + 1)x - (n + 2)x] = cos[nx + x - nx - 2x]
= cos[x - 2x] = cos(-x) = cosx[since cos x is an even function]
= RHS Hence proved.

LHS = (sin 3x + sin x) sin x + (cos 3x - cos x) cos x
= sin 3x sin x + sin²x + cos 3x cos x - cos²x
= sin 3x sin x + cos 3x cos x - (cos²x - sin²x)
= cos (3x - x) - cos 2x [∴ cosA cosB + sinA sinB = cos(A - B) and cos²A - sin²A = cos²A]
= 0 = RHS. Hence proved

RHS = tan 4x = tan2(2x)
= \(\frac { 2\quad tan\quad 2x }{ 1-{ tan }^{ 2 }2x } \left[ \because tan2x=\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right] \)
= \(\frac { 2.\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ 1-\left( \frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right) ^{ 2 } } =\frac { \frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ \frac { (1-{ tan }^{ 2 }x)^{ 2 }-4{ tan }^{ 2 }x }{ (1-{ tan }^{ 2 }x)^{ 2 } } } \)
= \(\frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }\quad x } \times \frac { (1-{ tan }^{ 2 }\quad { x })^{ 2 } }{ 1+{ tan }^{ 4 }x-2\quad { tan }^{ 2 }x-4{ tan }^{ 2 }x } \)
=  \(\frac { 4 tanx(1-{ tan }^{ 2 }\quad x) }{ 1+{ tan }^{ 4 }x-6\ { tan }^{ 2 }x } \)  = LHS