Question

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

Answer 1

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