Question

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

Answer 1

$\tan(2\alpha)=\dfrac{2\tan \alpha }{1-\tan^2 \alpha}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha}}{1-\dfrac{\sin ^2\alpha}{\cos^2\alpha}}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha} }{\dfrac{\cos^2\alpha}{\cos^2\alpha}-\dfrac{\sin ^2\alpha}{\cos^2\alpha}}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha} }{\dfrac{\cos^2\alpha-\sin^2 \alpha}{\cos^2\alpha}}$

$\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{2\sin\alpha}{\cos\alpha} \cdot\dfrac{\cos^2\alpha}{\cos^2\alpha-\sin^2 \alpha}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2 \alpha} \\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\sin(2\alpha)}{\cos(2\alpha)}$

Answer 2

Answer:

See below.

Step-by-step 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).