Question

The value of: $tan\left\{\left(sin\right)^{- 1} \left(cos \left(sin\right)^{- 1} x\right)\right\}\cdot tan\left\{\left(cos\right)^{- 1} \left(sin \left(cos\right)^{- 1} x\right)\right\};x\in \left(0 , \frac{\pi }{2}\right)$ is equal to:

Answer 1

Let
$L_{1}=\tan \left\{\sin ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right\} ; x \in\left(0, \frac{\pi}{2}\right)$ and $ L_{2}=\tan \left\{\cos ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right\} ; x \in\left(0, \frac{\pi}{2}\right) $ We have to find the value of $L_{1} \cdot L_{2}$ $ \begin{array}{l} L_{1}=\tan \left\{\sin ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right\} \Rightarrow L_{1}=\tan \left\{\sin ^{-1}\left(\cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)\right)\right\} \\ \Rightarrow L_{1}=\tan \left\{\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\right\} \\ \Rightarrow L_{1}=\tan \left\{\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right\} \\ \Rightarrow L_{1}=\frac{\sqrt{1-x^{2}}}{x} \end{array} $ Now take $ \begin{array}{l} L_{2}=\tan \left\{\cos ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right\} \\ \Rightarrow L_{2}=\tan \left\{\cos ^{-1}\left(\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)\right)\right\} \\ \Rightarrow L_{2}=\tan \left\{\cos ^{-1}\left(\sqrt{1-x^{2}}\right)\right\} \\ \Rightarrow L_{2}=\tan \left\{\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\right\} \\ \Rightarrow L_{2}=\frac{x}{\sqrt{1-x^{2}}} \\ \therefore L_{1} \cdot L_{2}=\frac{\sqrt{1-x^{2}}}{x} \times \frac{x}{\sqrt{1-x^{2}}}=1 \end{array} $