Inverse Trigono...

$${ sec\quad h }^{ -1 }\left( sin\quad \theta  \right) =$$

$$\cos^{ -1 }\left[\cos\left( -\frac { 17 }{ 15 } \pi  \right)  \right] $$ is equal to 

Find $$\displaystyle \int x.\sin xdx$$

If $$A=\tan^{1-}\left(\dfrac {x\sqrt {3}}{2k-x}\right)$$ and $$B=\tan^{-1}\left(\dfrac {2x-k}{k\sqrt {3}}\right)$$ then $$A.B=$$

If $$\cos^{-1}x-\cos^{-1}(\dfrac {y}{2})=\alpha$$ $$ax^{2}-4xy\cos \alpha +y^{2}=$$

if x$$>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }  \right)$$

$$\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 }  }  \right\} =\cos ^{ -1 }{ \dfrac { x }{ 2 } -\cos ^{ -1 }{ x }  }  }$$ holds for:

$$if\quad x>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }  \right) $$

If $$\cot^{-1}{x}+\tan^{-1}\left (\dfrac{1}{3}\right)=\dfrac{\pi}{2}$$, then $$x$$ will be

If $$x={ \sin }^{ -1 }(\sin10) $$ and $$y={ \cos }^{ -1 }(\cos10)$$, then find $$y - x$$.