If \(\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}\); then cos^-1 x + cos^-1 y is equal to

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

If the function f(x) = sin^-1(x² - 3), then x belongs to

The value of \({ cos }^{ -1 }\left( \cos\cfrac { 5\pi }{ 3 } \right) +sin^{ -1 }\left( \sin\cfrac{5\pi }{ 3 } \right) \) is ______________ 

If \({ tan }^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +{ tan }^{ -1 }\left( \frac { x-1 }{ x } \right) ={ tan }^{ -1 }\left( -7 \right) \) then x is ___________

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

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

Find all values of x such that -6\(\pi\le x \le 6\pi\) and cos x = 0 

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

Prove that tan (sin^-1x) = \(\frac{x}{\sqrt{1-x^{2}}} \), 1< x < 1

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

If \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1\) then find the value ofx.

Find the principal value of cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \)

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

Write the function \(f(x)=\tan ^{-1} \sqrt{\frac{a-x}{a+x}}-a<x<a \) in the simplest form

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