Question

Write the given expression in terms of x and y only tan(sin−1 x + cos−1 y)

Answer 1

Sinθ = x
cos α = y

 cos θ = square root of (1−x2) 
sin^2θ + cos^2θ = 1
cos θ = square root of (1- sin^2θ)
tan(sin−1 x + cos−1 y) = tan(θ + square root of (1- sin^2θ)) 
tan(sin−1 x + cos−1 y) = tan(θ+α)
tan(θ+α) =t an θ +tan α /(1-tanθ tanα)

Answer 2

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 $\tan \left ( \theta +\phi\right )=\frac{\tan \theta +\tan \phi }{1-\tan \theta \,\tan \phi }$

$\tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )=\frac{\frac{x}{\sqrt{1-x^2}}+\frac{\sqrt{1-y^2}}{y}}{1-\frac{x\sqrt{1-y^2}}{y\sqrt{-x^2}}}\\\therefore \tan \left ( \sin ^{-1}x+ \cos ^{-1}y\right )=\frac{xy+\sqrt{\left ( 1-x^2 \right )\left ( 1-y^2 \right )}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}$