Question

Prove:

$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

Answer 1

$LHS=\left(\frac{tan\theta }{1-cot\theta }\right)+\left(\frac{cot\theta }{1-tan\theta }\right)=\left(\frac{\left(\frac{sin\theta }{cos\theta }\right)}{1-\left(\frac{cos\theta }{sin\theta }\right)}\right)+\left(\frac{\left(\frac{cos\theta }{sin\theta }\right)}{1-\left(\frac{sin\theta }{cos\theta }\right)}\right)=\left(\frac{si{n}^2\theta }{cos\theta (sin\theta -cos\theta )}\right)+\left(\frac{co{s}^2\theta }{sin\theta (cos\theta -sin\theta )}\right)=\left(\frac{1}{sin\theta -cos\theta }\right)\left[\right(\frac{si{n}^2\theta }{cos\theta })-(\frac{co{s}^2\theta }{sin\theta }\left)\right]=\left(\frac{1}{sin\theta -cos\theta }\right)\left[\right(\frac{si{n}^3\theta -co{s}^3\theta }{sin\theta cos\theta }\left)\right]=\left(\frac{1\times (sin\theta -cos\theta )(si{n}^2\theta +sin\theta cos\theta +co{s}^2\theta )}{(sin\theta -cos\theta )sin\theta cos\theta }\right)=\left(\frac{1+sin\theta cos\theta }{sin\theta cos\theta }\right)=sec\theta \times cosec\theta +1=1+sec\theta cosec\theta =RHShenceproved!$