Question

Prove that $\frac{sin 2A}{(1 - cos 2A)} . \frac{(1 - cos A)}{cos A} = tan (\frac{A}{2})$

Answer 1

Answer:

$LHS=\frac{sin2A}{1-cos2A}.\frac{(1-cosA)}{cosA}\\ \\=\frac{2sinAcosA}{2sin^{2}A }.\frac{(1-cosA)}{cosA}\\ \\=\frac{cosA}{sinA}.\frac{(1-cosA)}{cosA}\\\\=\frac{1-cosA}{sinA}\\\\=\frac{2sin^{2}(A/2)}{2sin(A/2)cos(A/2)}\\\\=\frac{sin(A/2)}{cos(A/2)}\\\\=tan(\frac{A}{2} )\\\\=RHS$

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