Question

Prove that tan2A is equal to 2tanA/1-tan^2A

Answer 1

See below.

Explanation:

tan 2A = sin 2A / cos 2A

= 2 sinA cosA / (cos^2A - sin^2A)

Now divide top and bottom of the fraction by cos^2 A:

2 sinA cosA cos^2A sin^2 A

------------------- / ----------- - -------------

cos^2A cos^2 A cos^2 A

= 2 tan A / 1 - tan^2A).

Answer 2

`\tan(2\alpha)=(2\tan \alpha )/(1-\tan^2 \alpha)\\\\\\(\sin(2\alpha))/(\cos(2\alpha))=((2\sin\alpha)/(\cos\alpha))/(1-(\sin ^2\alpha)/(\cos^2\alpha))\\\\\\(\sin(2\alpha))/(\cos(2\alpha))=((2\sin\alpha)/(\cos\alpha) )/((\cos^2\alpha)/(\cos^2\alpha)-(\sin ^2\alpha)/(\cos^2\alpha))\\\\\\(\sin(2\alpha))/(\cos(2\alpha))=((2\sin\alpha)/(\cos\alpha) )/((\cos^2\alpha-\sin^2 \alpha)/(\cos^2\alpha))`

`(\sin(2\alpha))/(\cos(2\alpha))=(2\sin\alpha)/(\cos\alpha) \cdot(\cos^2\alpha)/(\cos^2\alpha-\sin^2 \alpha)\\\\\\(\sin(2\alpha))/(\cos(2\alpha))=(2\sin\alpha\cos\alpha)/(\cos^2\alpha-\sin^2 \alpha) \\\\\\(\sin(2\alpha))/(\cos(2\alpha))=(\sin(2\alpha))/(\cos(2\alpha))`