Question

Solve for x :
${\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta -\sin 2\theta =1+\tan \theta +\cot \theta$

Answer 1

Put $\sin 2\theta =2\sin \theta \cos \theta$

${\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta -2\sin \theta \cos \theta =1+\tan \theta +\cot \theta$
$(1-{\cos }^2\theta )\tan \theta +\frac{{\cos }^2\theta }{\tan \theta }-2\sin \theta \cos \theta =1+\tan \theta +\cot \theta$
$\tan \theta -{\cos }^2\theta \tan \theta +\frac{{\cos }^2\theta }{\tan \theta }-2\sin \theta \cos \theta =1+\tan \theta +\cot \theta$
$\tan \theta +\cot \theta -\sin 2\theta =1+\tan \theta +\cot \theta$
$\sin 2\theta =-1$
$2\theta =(4n-1)\frac{\pi }{2}$
$\theta =(4n-1)\frac{\pi }{4}$