Question

Tan2A - sin2A =sin2Atan2A​

Answer 1

Step-by-step explanation:

$\tan^2\alpha-\sin^2\alpha=\sin^2\alpha\tan^2\alpha\\\\LS=\left(\dfrac{\sin\alpha}{\cos\alpha}\right)^2-\sin^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}-\dfrac{\sin^2\alpha\cos^2\alpha}{\cos^2\alpha}\\\\=\dfrac{\sin^2\alpha-\sin^2\alpha\cos^2\alpha}{\cos^2\alpha}=\dfrac{\sin^2\alpha(1-\cos^2\alpha)}{\cos^2\alpha}=\dfrac{\sin^2\alpha\sin^2\alpha}{\cos^2\alpha}\\\\=\sin^2\alpha\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha}=\sin^2\alpha\left(\dfrac{\sin\alpha}{\cos\alpha}\right)^2=\sin^2\alpha\tan^2\alpha=RS$

Used:

$\tan x=\dfrac{\sin x}{\cos x}\\\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\sin^2\alpha+\cos^2\alpha=1\to \sin^2\alpha=1-\cos^2\alpha\\\\\text{distributive property}\ a(b+c)=ab+ac$