Answer to Question #162841 in Calculus for Aldryan venth

Question

Sec²(A/2)= 2-2 cos A/ sin² A

Answer 1

Use the trigonometric identity $1-cos\theta=2sin^2\theta$ to write the numerator as,

$2-2cosA=2(1-cosA)=4sin^2(\frac{A}{2})$

Use trigonometric identity $sin\theta=2sin({\frac{\theta}{2})}cos({\frac{\theta}{2})}$ to write the denominator as,

$sinA=2sin({\frac{A}{2})}cos({\frac{A}{2})}$

From Right-Hand side,

$\frac{2-2cosA}{sin^2A}=\frac{4sin^2(\frac{A}{2})}{(2sin\frac{A}{2}cos\frac{A}{2})^2}$

$=\frac{4sin^2(\frac{A}{2})}{4sin^2\frac{A}{2}cos^2\frac{A}{2}}$

The common term $4sin^2(\frac{A}{2})$ is cancel out from numerator and denominator.

$=\frac{1}{cos^2(\frac{A}{2})}$

$=sec^2(\frac{A}{2})=$ Left-Hand side

Therefore, $\frac{2-2cosA}{sin^2A}=$ $sec^2(\frac{A}{2})$ , hence proved.