Answer:
\(\tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )=\frac{xy+\sqrt{\left ( 1-x^2 \right )\left ( 1-y^2 \right )}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}\)
Step-by-step explanation:
We need to express \(\tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )\) in terms of x and y .
Let \(\sin ^{-1}x=\theta \,,\,\cos ^{-1}y=\phi\) , we get
\(x=\sin \theta \,,\,y=\cos \phi\)
Formulae Used:
\(\sin ^2\theta +\cos ^2\theta =1\\\sin ^2 \phi +\cos ^2 \phi =1\)
\(\cos \theta =\sqrt{1-\sin^2 \theta}=\sqrt{1-x^2}\\\sin \phi=\sqrt{1-\cos ^2 \phi}=\sqrt{1-y^2}\)
\(\tan \theta =\frac{\sin \theta }{\cos\theta }=\frac{x}{\sqrt{1-x^2}}\\\tan \phi =\frac{\sin \phi}{\cos\phi}=\frac{\sqrt{1-y^2}}{y}\)
We know that \(\tan \left ( \theta +\phi\right )=\frac{\tan \theta +\tan \phi }{1-\tan \theta \,\tan \phi }\)
\(\tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )=\frac{\frac{x}{\sqrt{1-x^2}}+\frac{\sqrt{1-y^2}}{y}}{1-\frac{x\sqrt{1-y^2}}{y\sqrt{-x^2}}}\\\therefore \tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )=\frac{xy+\sqrt{\left ( 1-x^2 \right )\left ( 1-y^2 \right )}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}\)
