Answer
Ans. Proved LHS=RHS
$tanθ/2=\frac{sinθ}{1+cosθ}$
Work Step by Step
Given:-$tanθ/2=\frac{sinθ}{1+cosθ}$
Taking RHS
=$\frac{sinθ}{1+cosθ}$
By using the following identities $sinθ=2sin{θ/2}cos{θ/2}, cosθ=cos^2{θ/2}-sin^2{θ/2}, cos^2{θ/2}+sin^2{θ/2}=1$
=$\frac{2sin{θ/2}cos{θ/2}}{1+cos^2{θ/2}-sin^2{θ/2}}$
=$\frac{2sin{θ/2}cos{θ/2}}{cos^2{θ/2}+sin^2{θ/2}+cos^2{θ/2}-sin^2{θ/2}}$
=$\frac{2sin{θ/2}cos{θ/2}}{2cos^2{θ/2}}$
=$tanθ/2$
Ans.
