Question

Prove $\frac{3-tan\theta }{3cosec\theta -sec\theta }$$=\frac{\sin \theta \cos \theta +\sin 2\theta -{\sin }^2\theta }{(\cos \theta -\sin \theta )}$

Answer 1

Consider the given expression.

$\frac{3-\tan \theta }{3\mathrm{cosec}\theta -\sec \theta }$

$=\frac{3-\frac{\sin \theta }{\cos \theta }}{3\times \frac{1}{\sin \theta }-\frac{1}{\cos \theta }}$

$=\frac{\frac{3\cos \theta -\sin \theta }{\cos \theta }}{\frac{\cos \theta -\sin \theta }{\sin \theta \cos \theta }}$

$=\frac{(3\cos \theta -\sin \theta )\sin \theta }{(\cos \theta -\sin \theta )}$

$=\frac{(3\sin \theta \cos \theta -{\sin }^2\theta )}{(\cos \theta -\sin \theta )}$

$=\frac{\sin \theta \cos \theta +2\sin \theta \cos \theta -{\sin }^2\theta }{(\cos \theta -\sin \theta )}$

$=\frac{\sin \theta \cos \theta +\sin 2\theta -{\sin }^2\theta }{(\cos \theta -\sin \theta )}$

Hence, Proved