Question

Prove that 2sin theta /1+cos theta+simtheta=1-costheta+sintheta/1+sintheta

Answer 1

Start with the R.H.S:

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

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

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

Therefore, L.H.S = R.H.S

Answer 2

Solution:

LHS
Formula used:
$a^{2}-b^{2}=(a+b)(a-b)\\\\1 - {cos}^{2}\theta= {sin}^{2}\theta$

$\frac{2 \: sin \: \theta}{1 + cos \: \theta + sin \: \theta} \times \frac{1 - cos \: \theta + sin \: \theta}{1 - cos \: \theta + sin \: \theta} \\ \\ = \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{(1 + cos \: \theta + sin \: \theta)(1 - cos \: \theta + sin \: \theta)} \\ \\ \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{(1 - {cos}^{2}\theta + sin \: \theta - sin \: \theta \: cos \: \theta + sin \: \theta + sin \: \theta \: cos \: \theta + {sin}^{2}\theta )} \\ \\ = \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{(1 - {cos}^{2}\theta + sin \: \theta + sin \: \theta+ {sin}^{2}\theta )} \\ \\ = \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{({sin}^{2}\theta + sin \: \theta + sin \: \theta+ {sin}^{2}\theta )} \\ \\ = \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{(2{sin}^{2}\theta + 2sin \: \theta )} \\ \\ = \frac{2\: sin \: \theta(1 - cos \: \theta + sin \: \theta)}{2 \: sin \: \theta(1 + sin \: \theta )} \\ \\ = \frac{1 - cos \: \theta + sin \: \theta}{1 + sin \: \theta} \\ \\ = R.H.S.$
Hope it helps you.