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

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 $\begin{array}{l}{L}_2=\tan \left\{{\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)\right\}\\ \Rightarrow L_2=\tan \left\{{\cos }^{-1}\left(\sin \left({\sin }^{-1}\sqrt{1-x^2}\right)\right)\right\}\\ \Rightarrow L_2=\tan \left\{{\cos }^{-1}\left(\sqrt{1-x^2}\right)\right\}\\ \Rightarrow L_2=\tan \left\{{\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right\}\\ \Rightarrow L_2=\frac{x}{\sqrt{1-x^2}}\\ \therefore L_1\cdot L_2=\frac{\sqrt{1-x^2}}{x}\times \frac{x}{\sqrt{1-x^2}}=1\end{array}$