Question

Prove that:

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

Answer 1

Lets take LHS and then equate it to RHS.

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

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

$=\frac{(1+\cos \theta +\sin \theta {)}^2}{(1+\cos \theta {)}^2-{\sin }^2\theta }$ [$\because (a+b)(a-b)=a^2-b^2$]

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

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

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

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

$=\frac{2\left(\right(1+\cos \theta )+\sin \theta (1+\cos \theta \left)\right)}{2\left((1+\cos \theta )\cos \theta \right)}$

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

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

$=RHS$

Therefore, $\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }=\frac{1+\sin \theta }{\cos \theta }$

Hence proved