# PROVE THAT : [sinA /1-cosA - 1-cosA/sinA ][cosA/1+sinA + 1+sinA/cosA] = 4cosecA.?

Oct 24, 2016

$L H S = \left[\sin \frac{A}{1 - \cos A} - \frac{1 - \cos A}{\sin} A\right] \left[\cos \frac{A}{1 + \sin A} + \frac{1 + \sin A}{\cos} A\right]$

Now 1st part= $\sin \frac{A}{1 - \cos A} - \frac{1 - \cos A}{\sin} A$

$= \frac{\sin A \left(1 + \cos A\right)}{\left(1 - \cos A\right) \left(1 + \cos A\right)} - \frac{1 - \cos A}{\sin} A$

=(sinA(1+cosA))/(1-cos^2A) -(1/sinA-cosA/sinA)

$= \frac{\sin A \left(1 + \cos A\right)}{\sin} ^ 2 A - \csc A + \cot A$

$= \frac{1 + \cos A}{\sin} A - \csc A + \cot A$

$= \frac{1}{\sin} A + \cos \frac{A}{\sin} A - \csc A + \cot A$

$= \csc A + \cot A - \csc A + \cot A$

$= 2 \cot A$

2nd part

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

$= \frac{\cos A \left(1 - \sin A\right)}{\left(1 + \sin A\right) \left(1 - \sin A\right)} + \left(\frac{1}{\cos} A + \sin \frac{A}{\cos} A\right)$

$= \frac{\cos A \left(1 - \sin A\right)}{1 - {\sin}^{2} A} + \left(\sec A + \tan A\right)$

$= \frac{\cos A \left(1 - \sin A\right)}{\cos} ^ 2 A + \left(\sec A + \tan A\right)$

$= \frac{1 - \sin A}{\cos} A + \left(\sec A + \tan A\right)$

$= \frac{1}{\cos} A - \sin \frac{A}{\cos} A + \sec A + \tan A$

$= \sec A - \tan A + \sec A + \tan A$

$= 2 \sec A$

Whole LHS

$= 2 \cot A \times 2 \sec A$

$= \frac{4 \cos A}{\sin} A \times \frac{1}{\cos} A$

$= \frac{4}{\sin} A = 4 \csc A = R H S$

Proved