Question

[(1+sin theta)÷(1+cos theta)]+[(1-sin theta)÷(1-cos theta)]=2(cosec^2 theta-cot theta)

Answer 1

hope it helpful for u....
Thx...

Answer 2

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

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

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

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

$\boxed {\pink { Since, 1 - cos^{2} \theta = sin^{2} \theta }}$

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

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

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

$\green {= 2( Cosec^{2} \theta - cot \theta )} \\= RHS$

•••♪