LHS = \({ cot }^{ -1 }\left[ \frac { \sqrt { 1+sinx } +\sqrt { 1-sinx } }{ \sqrt { 1+sinx } -\sqrt { 1-sinx } } \right] \)
=  \({ cot }^{ -1 }\left[ \frac { \sqrt { 1+sinx } +\sqrt { 1-sinx } }{ \sqrt { 1+sinx } -\sqrt { 1-sinx } } \times \frac { \sqrt { 1+sinx } +\sqrt { 1+sinx } }{ \sqrt { 1+sinx } +\sqrt { 1+sinx } } \right] \)    [multiplying the numerator and denominator by the conjugate of the denominator]
=  \({ cot }^{ -1 }\left[ \frac { \left( \sqrt { 1+sinx } +\sqrt { 1-sinx } \right) ^{ 2 } }{ (\sqrt { 1+sinx } )^{ 2 }-(\sqrt { 1-sinx } )^{ 2 } } \right] \)

= \({ cot }^{ -1 }\left[ \frac { 1+sinx+1-sinx+2\sqrt { (1+sin\quad x)(1-sin\quad x) } }{ (1+sin\quad x)-(1-sin\quad x) } \right] \)
=  \({ cot }^{ -1 }\left[ \frac { 2+2\sqrt { 1-{ sin }^{ 2 }x } }{ 2\quad sin\quad x } \right] ={ cot }^{ -1 }\left[ \frac { 2+2\quad cos\quad x }{ sin\quad x } \right] \left[ \because 1-{ sin }^{ 2 }x={ cos }^{ 2 }x \right] \)
= \({ cot }^{ -1 }\left[ \frac { 1+cos\quad x }{ sin\quad x } \right] ={ cot }^{ -1 }\left[ \frac { 2{ cos }^{ 2 }\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } } \right] \)    \(\left[ \because sin\quad A=2 \ \sin\frac { A }{ 2 } cos\frac { A }{ 2 } \quad and\quad 1+cos\quad x=2{ cos }^{ 2 }\frac { x }{ 2 } \right] \)
= \({ cot }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } }{ sin\frac { x }{ 2 } } \right] ={ cot }^{ -1 }\left( cot\frac { x }{ 2 } \right) \)
= \(\frac { x }{ 2 } \) =RHS
Hence proved.

Given \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)={ sin }^{ -1 }(1)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =\frac { \pi }{ 2 } -{ cos }^{ -1 }(x)\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)  \(\left[ \because { sin }^{ -1 }(x)+{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \right] \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)
⇒ \(x=\frac { 1 }{ 5 } \)
Thus, \(x=\frac { 1 }{ 5 } \) is a root of the given equation.

LHS = cos 6x
= cos 3(2x) = 4 cos³2x- 3cos2x    [∴ cos 3\(\theta\) = 4 cos³\(\theta\) - 3 cos\(\theta\)]
= 4 (2cos²x - 1)³- 3(2cos²x - 1)  [∴ cos2x = 2cos²x - 1]
= 4 (8 cos⁶x-12 cos⁴x + 6 cos²x-1) - 6 cos²x + 3
= 32 cos⁶x - 48 cos⁴x + 18 cos²x - 1
= RHS. Hence proved.

LHS = cos 4x = cos 2(2x)
= 2 cos²2x-1 [∴ cos 2A = 2cos²A - 1]
= 2 [2 cos²x - 1]² - 1 = 2 [4 cos⁴x - 4cos2x + 1] - 1
= 8 cos⁴x - 8 cos²x + 2 - 1 = 1 - 8 cos²x + 8 cos⁴x
= 1 - 8 cos²x (1 - cos²x) = 1 - 8 cos²x . sin²x = RHS.

LHS = cot x cot 2x - cot 2 x cot 3 x - cot 3x cot x
We have 3x = x + 2.x
\(cot\quad 3x=\frac { cotx\quad cot2x-1 }{ cotx+cot2x } \)   \(\left[ \therefore tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB } \right] \)
cross multiplying we get,
cot x cot 3x + cot 2x cot 3x = -1 + cot x cot 2x
\(\Rightarrow\) cot x cot 2.x - cot 2x cot 3x - cot x cot 3x = 1
Hence proved