Question

$\frac{\sin \theta }{1-\cos \theta }+\frac{\tan \theta }{1+\cos \theta }=\sec \theta \text{cosec}\theta +\cot \theta$

Answer 1

To prove: $\frac{\sin \theta }{1-\cos \theta }+\frac{\tan \theta }{1+\cos \theta }=\sec \theta \text{cosec}\theta +\cot \theta$

LHS $=\frac{\sin \theta }{1-\cos \theta }+\frac{\tan \theta }{1+\cos \theta }$

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

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

$=\frac{\sin \theta +\sin \theta \cos \theta +\frac{\sin \theta }{\cos \theta }-\sin \theta }{{\sin }^2\theta }$

$=\frac{\sin \theta \cos \theta +\frac{\sin \theta }{\cos \theta }}{{\sin }^2\theta }$

$=\frac{\sin \theta \cos \theta }{{\sin }^2\theta }+\frac{\sin \theta }{\cos \theta }\cdot \frac{1}{{\sin }^2\theta }$

$=\frac{\cos \theta }{\sin \theta }+\frac{1}{\cos \theta }\frac{1}{\sin \theta }$

$=\cot \theta +\sec \theta \text{cosec}\theta$

$=$ RHS

$\therefore \frac{\sin \theta }{1-\cos \theta }+\frac{\tan \theta }{1+\cos \theta }=\sec \theta \text{cosec}\theta +\cot \theta$

Hence, proved.