${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

$\tan \left[ \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right) \right] = $

माना $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\sin x} \right) = l$ और $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\tan x} \right) = m$,तो

$\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=$ . . . . . . .

यदि $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$ है,तो $\sin^{-1}\left(\frac{A}{C}\right) = $

$\cos \left( \sin^{-1} \frac{5}{13} \right) = $