Question

prove that sin2A/(1+cos2A)×cosA/(1+cosA)=tanA/2

Answer 1

please mark it brainleist

Answer 2

Answer:

To prove: $\frac{sin\,2A}{1+cos\,2A}\times\frac{cos\,A}{1+cos\,A}=tan\,\frac{A}{2}$

using,

cos 2x = 2.cos²x - 1

⇒ 1 + cos 2x = 2.cos²x

and sin 2x = 2 . sin x . cos x

and $sin\,x=\frac{2.tan\,\frac{x}{2}}{1+tan^2\,\frac{x}{2}}$

and $sin\,x=\frac{1-tan^2\,\frac{x}{2}}{1+tan^2\,\frac{x}{2}}$

Now, Consider

LHS

$=\frac{sin\,2A}{1+cos\,2A}\times\frac{cos\,A}{1+cos\,A}$

$=\frac{2.sin\,A.cos\,A}{2.cos^2\,A}\times\frac{cos\,A}{1+cos\,A}$

$=\frac{sin\,A}{1+cos\,A}$

$=\frac{\frac{2.tan\,\frac{A}{2}}{1+tan^2\,\frac{A}{2}}}{1+\frac{1-tan^2\,\frac{A}{2}}{1+tan^2\,\frac{A}{2}}}$

$=\frac{\frac{2.tan\,\frac{A}{2}}{1+tan^2\,\frac{A}{2}}}{\frac{1+tan^2\,\frac{A}{2}+1-tan^2\,\frac{A}{2}}{1+tan^2\,\frac{A}{2}}}$

$=\frac{2.tan\,\frac{A}{2}}{1+tan^2\,\frac{A}{2}+1-tan^2\,\frac{A}{2}}$

$=\frac{2.tan\,\frac{A}{2}}{2}$

$=tan\,\frac{A}{2}$

$=RHS$

Hence Proved.