Question

Prove that:
$\frac{2\cos 62\theta -1}{\sin \theta \cos \theta }=\cot \theta -\tan \theta$

Answer 1

$\frac{2co{s}^2\theta -1}{sin\theta .cos\theta }=cot\theta -tan\theta$

R.H.S :- $cot\theta -tan\theta$

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

$\frac{co{s}^2\theta -si{n}^2\theta }{cos\theta sin\theta }$

$\frac{co{s}^2\theta -(1-co{s}^2\theta )}{cos\theta sin\theta }$

$\frac{co{s}^2\theta -1+co{s}^2\theta }{cos\theta .sin\theta }.$

$=\frac{2co{s}^2\theta -1}{sin\theta .cos\theta }.$

$=L.H.S.$