Question

If $\sin \theta -\sqrt{6}\cos \theta =\sqrt{7}\cos \theta$. Prove that $\cos \theta -\sqrt{6}\sin \theta -\sqrt{7}\sin \theta =1$.

Answer 1

$\to \sin \theta -\sqrt{6}\cos \theta =\sqrt{7}\cos \theta$

$\to \tan \theta =\sqrt{6}+\sqrt{7}$

$\to \cot \theta (\sqrt{6}+\sqrt{7})=1$

$\to \sqrt{6}\sin \theta +\sqrt{7}\sin \theta =\cos \theta$

$\to \cos \theta -\sqrt{6}\sin \theta -\sqrt{7}\sin \theta =1$