Question

Tan2A - sin2A =sin2Atan2A​

Answer 1

Explanation:

`\tan^2\alpha-\sin^2\alpha=\sin^2\alpha\tan^2\alpha\\\\LS=\left((\sin\alpha)/(\cos\alpha)\right)^2-\sin^2\alpha=(\sin^2\alpha)/(\cos^2\alpha)-(\sin^2\alpha\cos^2\alpha)/(\cos^2\alpha)\\\\=(\sin^2\alpha-\sin^2\alpha\cos^2\alpha)/(\cos^2\alpha)=(\sin^2\alpha(1-\cos^2\alpha))/(\cos^2\alpha)=(\sin^2\alpha\sin^2\alpha)/(\cos^2\alpha)\\\\=\sin^2\alpha\cdot(\sin^2\alpha)/(\cos^2\alpha)=\sin^2\alpha\left((\sin\alpha)/(\cos\alpha)\right)^2=\sin^2\alpha\tan^2\alpha=RS`

Used:

`\tan x=(\sin x)/(\cos x)\\\\\left((a)/(b)\right)^n=(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`