Question

Prove that
$\frac{tan\theta }{1-\cot \theta }+\frac{cot\theta }{1-tan\theta }=1+\sec \theta cos\theta$

Answer 1

$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }\frac{\tan \theta }{1-\cot \theta }-\frac{{\cot }^2\theta }{1-\cot \theta }=\frac{\tan \theta -{\cot }^2\theta }{1-\cot \theta }=\frac{1-{\cot }^3\theta }{\cot \theta (1-\cot \theta )}=\frac{1+\cot \theta +{\cot }^2\theta }{\cot \theta }=\tan \theta +1+\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }+1=\sec \theta \csc \theta +1$