**Toolbox:**

- Put $ x = sin \theta\: and \: y = sin\phi$
- Simplify and differentiate after back substituting $ \theta\: and \: \phi$
- $ cosx+cosy=2cos\large\frac{x+y}{2}.cos\large\frac{x-y}{2}$
- $ sinx-siny=2cos\large\frac{x+y}{2}.sin\large\frac{x-y}{2}$

$ cos\theta+cos\phi = a(sin\theta-sin\phi)$

$ 2cos\large\frac{\theta+\phi}{2}\: cos\large\frac{\theta-\phi}{2}=2a\: cos\large\frac{\theta+\phi}{2}\: sin\large\frac{\theta-\phi}{2}$

$ \theta-\phi=2tan^{-1}a$

$ tan\large\frac{\theta-\phi}{2}=a$

$ \Rightarrow 2tan^{-1}a=\theta-\phi$

\( \Rightarrow sin^{-1}x-sin^{-1}y=2tan^{-1}a\)

Differentiate w.r. to \( x \) on both sides to get the result.

\( \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dy}=0\)

\( \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}\)