Question

Differentiate ${\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$.

Answer 1

$d(a\cos x-b\sin x)=-a\sin x-b\sin x=-(a\sin x+b\cos x)$

$d(b\cos x+a\sin x)=-b\sin x+a\cos x$

$d(\frac{a\cos x-b\sin x}{b\cos x+a\sin x})=dy=-\frac{(a\sin x+b\cos x{)}^2+(a\cos x-b\sin x{)}^2}{(b\cos x+a\sin x{)}^2}=-(1+y^2)$ where $y=\frac{a\cos x-b\sin x}{b\cos x+a\sin x}$

$d\left(ta{n}^{-1}y\right)=\frac{1}{1+y^2}dy$

$⟹-1$