Inverse Trigonometric Functions

If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x² $$-$$ 1 is :

The value of $$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$ is :

The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :

Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$. Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :

The set of all values of k for which $${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$$, is the interval :

Let m and M respectively be the minimum and the maximum values of $$f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$$. Then m + M is equal to :

$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$$ is equal to :

If $$0 < x < {1 \over {\sqrt 2 }}$$ and $${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }$$, then the value of $$\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)$$ is :

The sum of the absolute maximum and absolute minimum values of the function $$f(x)=\tan ^{-1}(\sin x-\cos x)$$ in the interval $$[0, \pi]$$ is :

Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$. If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :