\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right) \)
\(\text { Put } x^{2}=\cos 2 \theta, \text { then } \)
\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\) 
\( \tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right)\) 
\(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)
[divide numerator and denominator inside the bracket by \(cos \theta]\)

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

\(\text {We have } 2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x) \)
\(\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}\left(\frac{2}{\cos x}\right) \)
\(\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=\frac{2}{\cos x} \Rightarrow \frac{\sin x}{\cos x}=1 \)
\(\Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \)