Inverse Trigono...

Find the values of $$\cos^{-1}\left(\cos \dfrac {7\pi}{6}\right)$$ is equal to

The value of $$3\tan^{-1}\dfrac {1}{2}+2\tan^{-1}\dfrac {1}{5}+\sin^{-1}\dfrac {142}{65\sqrt {5}}$$ is

Let in $$\Delta ABC, \, \angle A = \dfrac{\pi}{2}$$. Then value of $$\tan^{-1} \dfrac{b}{a + c} + \tan^{-1} \dfrac{c}{a + b}$$ equals 

Find the approximate value of $$\sin^{-1}(0.51)$$ given that $$\sqrt {3}=1.7321$$

$${\sin ^{ - 1}}(\cos \left( {{{\sin }^{ - 1}}x} \right)) + {\cos ^{ - 1}}(\sin \left( {{{\cos }^{ - 1}}x} \right)){\text{is}}\;{\text{equal}}\;{\text{to}}{\text{.}}$$

$$\cot ^{ -1 }{ \left( \dfrac { \sqrt { 1-\sin { x }  } +\sqrt { 1+\sin { x }  }  }{ \sqrt { 1-\sin { x }  } -\sqrt { 1+\sin { x }  }  }  \right)  }$$=....$$\left( 0<x<\dfrac { \pi  }{ 2 }  \right)$$

Find the value of $${\sin ^{ - 1}}\left( 1 \right)$$

If $${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{1}{x} + {\tan ^{ - 1}}\dfrac{1}{{{a^2} - x + 1}},\,then\,x\,is\,$$

$${\cot ^{ - 1}}\left( {2 + \sqrt 3 } \right) = $$

$$cot^{-1}(\sqrt{cos\alpha })-tan^{-1}(\sqrt{cos\alpha })=x$$ , then sin x is equal to