$\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$
I've tried and cannot prove it
Hi,
This equality can be written $\displaystyle \sin^2x-\tan x =\tan^2x\left(\cos^2x-\cot x\right)$. You should try to use $\displaystyle \begin{cases}\tan x = \frac{\sin x }{\cos x}\\ \cot x =\frac{1}{\tan x}\end{cases}$ to show that $\displaystyle \tan^2x\left(\cos^2x-\cot x\right)$ equals $\displaystyle \sin^2x-\tan x$.
$\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$
Working with Left Hand Side:
$\displaystyle \frac{\sin^2 x-\frac{\sin x}{\cos x}}{\cos^2 x-\frac{\cos x}{\sin x}}$
$\displaystyle \frac {\frac{\sin^2 x \cos x - \sin x}{\cos x}}{\frac {\sin x \cos^2 x-\cos x}{\sin x}}$
$\displaystyle \frac {\frac{\sin x(\sin x \cos x - 1)}{\cos x}}{\frac {\cos x(\sin x \cos x-1)}{\sin x}}$
$\displaystyle \frac{\sin x(\sin x \cos x - 1)}{\cos x} \cdot \frac {\sin x}{\cos x(\sin x \cos x-1)}$
$\displaystyle \frac{\sin^2 x}{\cos^2 x}$
$\displaystyle \tan^2 x$