(2sinA + sin2A)/(2sinA - sin2A) = cot²A/2 | Prove | Trigonometry

Prove the following trigonometric identity: $\dfrac{2sinA + sin2A}{2sinA - sin2A} = cot^2 \frac{A}{2}$

Solution:

LHS

$= \dfrac{2sinA + sin2A}{2sinA - sin2A}$

[Using sin2A = 2sinAcosA]

$= \dfrac{2sinA + 2sinAcosA}{2sinA - 2sinAcosA}$

$= \dfrac{2sinA(1+cosA)}{2sinA(1-cosA)}$

$= \dfrac{1 + cosA}{1-cosA}$

[Using sub-multiple angle formula of cosA]

$= \dfrac{1 + (2cos^2\frac{A}{2} - 1)}{1-(1-2sin^2\frac{A}{2})}$

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

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

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

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

RHS

#proved