Hello Hellooo
Welcome to Math Help Forum! Originally Posted by
Hellooo How someone help me prove the identity below?
(1-tan^2X)(1-cos^2X)=[(sin^2X-4sin^4X)/(1-sin^2X)]
Thanks in advance! =)
I think the 4 on the RHS should be a 2. Here's my answer:
$\displaystyle (1-\tan^2x)(1-\cos^2x)$$\displaystyle =\left(1-\frac{\sin^2x}{\cos^2x}\right)\sin^2x$, using $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and $\displaystyle 1 - \cos^2x = \sin^2 x$
$\displaystyle =\frac{(\cos^2x-\sin^2x)\sin^2x}{\cos^2x}$
$\displaystyle =\frac{(1-2\sin^2x)\sin^2x}{1-\sin^2x}$, using $\displaystyle \cos^2x =1-\sin^2x$
$\displaystyle =\frac{\sin^2x-2\sin^4x}{1-\sin^2x}$
Grandad