Question

Please solve it.........​

Answer 1

Answer:

Please refer the image

Step-by-step explanation:

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.