Question

Sovle ${\sin }^{2}x\tan x+{\cos }^{2}x\cot x+\sin 2x=\tan x+\cot x$.

Answer 1

$LHS={\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta +2\sin \theta \cos \theta$

$={\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta +2\sin \theta \cos \theta$

$=(1-{\cos }^2\theta )\tan \theta +(1-{\sin }^2\theta )\cot \theta +2\sin \theta \cos \theta$

$=\tan \theta +\cot \theta -{\cos }^2\theta \tan \theta -{\sin }^2\theta \cos \theta +2\sin \theta \cos \theta$

$=\tan \theta +\cot \theta -\sin \theta \cos \theta -\sin \theta \cos \theta +2\sin \theta \cos \theta$

$=\tan \theta +\cot \theta$

$=RHS$