Question

${\tan }^{-1}\left[\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right]$= ${\tan }^{-1}\left(\frac{a}{b}\right)-x$

• A. True
• B. False

Answer 1

The correct option is A True

$ta{n}^{-1}\left[\right(\frac{acosx-bsinx}{bcosx+asinx}\left)\right]=ta{n}^{-1}\left[\right(\frac{a-btanx}{b+atanx}\left)\right]=ta{n}^{-1}\left[\right(\frac{\left(\frac{a}{b}\right)-tanx}{1+\left(\frac{a}{b}\right)tanx}\left)\right]let\left(\frac{a}{b}\right)=tan\theta then=ta{n}^{-1}\left[\right(\frac{tan\theta -tanx}{1+tan\theta tanx}\left)\right]=ta{n}^{-1}\left[tan\right(\theta -x\left)\right]=\theta -x=ta{n}^{-1}\left(\frac{a}{b}\right)-x$