\(\sin ^{-1}(\cos x)=\frac{\pi}{2}-x\) is valid for

If \(\cot ^{-1} x=\frac{2 \pi}{5}\) for some x \(\in\) R, the value of tan^-1 x is

The number of solutions of the equation \({ tan }^{ -1 }2x+{ tan }^{ -1 }3x=\frac { \pi }{ 4 } \) ____________

\({ tan }^{ -1 }\left( \frac { 1 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \) = ____________

If \({ cos }^{ -1 }x>x>{ sin }^{ -1 }x\) then _________

For what value of x does sinx = sin⁻¹x?

Find the value of sin^-1\(\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \).

Show that cot⁻¹\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

Prove that \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) ={ tan }^{ -1 }\left( \frac { 2 }{ 9 } \right) \)

Evaluate \(sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) \)

For what value of x, the inequality \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \) holds?

Find the value of
 \(cos\left( { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)

Solve: \({ tan }^{ -1 }\left( \cfrac { x-1 }{ x-2 } \right) +{ tan }^{ -1 }\left( \cfrac { x+1 }{ x+2 } \right) =\cfrac { \pi }{ 4 } \)

Solve \({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) +{ cot }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 2x } \right) =\frac { \pi }{ 3 } ,x>0\)

Solve: cos(tan^-1x) = \(sin\left( { cot }^{ -1 }\frac { 3 }{ 4 } \right) \) 

Solve \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)

Solve \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =sin\left\{ cot^{ -1 }\left( \frac { 3 }{ 4 } \right) \right\} \)

If \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\) than prove that x² = sin 2a

Prove that \({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) ={ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+{ x }^{ 2 } } .\sqrt { 1+{ y }^{ 2 } } } \right)\)