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