Question:Prove that: $\dfrac{1 + cos \alpha}{1 - cos \alpha} = cot^2 \frac{ \alpha}{2}$

Solution:

Taking LHS

$= \dfrac{1 + cos \alpha}{1 - cos \alpha}$


Using sub-multiple formula of cosA and trigonometric identity of 1.

$= \dfrac{(sin^2 \frac{\alpha}{2} + cos^2 \frac{\alpha}{2}) + (cos^2 \frac{\alpha}{2} - sin^2 \frac{\alpha}{2})}{(sin^2 \frac{\alpha}{2} + cos^2 \frac{\alpha}{2}) - (cos^2 \frac{\alpha}{2} - sin^2 \frac{\alpha}{2})$

$= \dfrac{2cos^2 \frac{\alpha}{2}}{2 sin^2 \frac{\alpha}{2}}$

$= cot^2 \frac{\alpha}{2}$

RHS

#proved