Question

$\frac{1+sin\theta }{1+cos\theta }+\frac{1-sin\theta }{1-cos\theta }=2.(cose{c}^2\theta -cot\theta )$

Answer 1

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

To be $LHS=\frac{1-\cos \theta +\sin \theta -\sin \theta .\cos \theta +1-\sin \theta +\cos \theta -\sin \theta -\cos \theta }{(1+\cos \theta )(1-\cos \theta )}$

$=\frac{2-2\sin \theta .\cos \theta }{(1-{\cos }^2\theta )}$

$=2\frac{(1-\sin \theta .\cos \theta )}{{\sin }^2\theta }$ divide By ${\sin }^{2}e$

$=2({\csc }^2\theta -\cot \theta )$

Hence proved