Question

If $x \cos \theta+y \sin \theta=\cos (\theta+\alpha)$ and $x^2+y^2=1$, then $\alpha$ is equal to

(A) $\sin ^{-1} y$

(B) $-\sin ^{-1} y$

(C) $\tan ^{-1} \frac{x}{y}$

(D) $-\tan ^{-1}\left(\frac{x}{y}\right)$

Answer 1

$\because x \cos \theta+y \sin \theta=\cos \theta \cos \alpha-\sin \theta \sin \alpha$

On comparing $\cos \theta$ and $\sin \theta$, we get

$x=\cos \alpha$ and $y=-\sin \alpha$

$\therefore x^2+y^2 =\cos ^2 \alpha+(-\sin \alpha)^2=1$

$y =-\sin \alpha$

$\Rightarrow \alpha  =\sin ^{-1}(-y)=-\sin ^{-1}(y)$

So, The correct option will be (B).