Typical trigonometry question for IIT JEE

$b\sin x+a\cos x =a$

Divide both sides by $\sqrt{a^2+b^2}$

$\dfrac{b}{\sqrt{a^2+b^2}}\sin x+\dfrac{a}{\sqrt{a^2+b^2}} \cos x=\dfrac{a}{\sqrt{a^2+b^2}}$

Take $\dfrac{b}{\sqrt{a^2+b^2}}=\sin \alpha \Rightarrow \dfrac{a}{\sqrt{a^2+b^2}}=\cos \alpha\Rightarrow \tan\alpha =\dfrac{b}{a}$

$\Rightarrow \sin\alpha \sin x+\cos\alpha \cos x=\cos\alpha$

$\Rightarrow \cos(x-\alpha)=\cos\alpha$

$\Rightarrow x-\alpha=2n\pi \pm \alpha$

$\Rightarrow x=2n\pi + 2\alpha$ or $x = 2n\pi$ (Told in the question to reject)

Hence $\sin x+\cos x =\sin(2n\pi+2\alpha)+\cos (2n\pi+2\alpha) =\sin 2\alpha+\cos 2\alpha$

$= \dfrac{2\tan \alpha}{1+\tan^2\alpha}+\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}=\dfrac{2ab+a^2-b^2}{a^2+b^2}$