Inverse Trigono...

$$Sin h (cos h ^{-1} x) =$$

The value of $$e^{sinh^{-1} (tan \theta)}$$ is equal to

$$\tanh^{1}\left(\dfrac {1}{3}\right)+\coth^{1}(3)=$$..... 

For which value of x, $$\sin [\cot^{-1}(x+1)]=\cos (\tan^{-1}x)$$.

If $$cot^{-1} x - cot^{-1} (x+2) = 15^0$$ then x is equal to 

If $$y=\dfrac{1}{2}\csc\ h^{-1}{\left(\dfrac{1}{2x\sqrt{1+x^{2}}}\right)}$$ then $$x$$=

Solve : $$sin h^{-1} (\dfrac{x}{\sqrt{1-x^2}}) =$$

The value of $$\tan^{-1}\left[\dfrac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right],\left|x\right|<\dfrac{1}{2},x=0$$, is equal to:

The value of $$\tan{\left\{\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}+\tan{\left\{\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}$$

If $$f(x)={ sin }^{ -1 }\left( \frac { \sqrt { 3 }  }{ 2 } x-\frac { 1 }{ 2 } \sqrt { 1-{ x }^{ 2 } }  \right) -\frac { 1 }{ 2 } \le x\le 1$$, then f(x) is equal to :