# Prove Trig Identity

• Nov 8th 2008, 11:52 AM
Raj
Prove Trig Identity
$\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

I've tried and cannot prove it (Headbang)
• Nov 8th 2008, 12:26 PM
flyingsquirrel
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 $\displaystyle \sin^2x-\tan x$.
• Nov 8th 2008, 12:26 PM
masters
Quote:

Originally Posted by Raj
$\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

I've tried and cannot solve it (Headbang)

$\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

Working with Left Hand Side:

$\displaystyle \frac{\sin^2 x-\frac{\sin x}{\cos x}}{\cos^2 x-\frac{\cos x}{\sin x}}$

$\displaystyle \frac {\frac{\sin^2 x \cos x - \sin x}{\cos x}}{\frac {\sin x \cos^2 x-\cos x}{\sin x}}$

$\displaystyle \frac {\frac{\sin x(\sin x \cos x - 1)}{\cos x}}{\frac {\cos x(\sin x \cos x-1)}{\sin x}}$

$\displaystyle \frac{\sin x(\sin x \cos x - 1)}{\cos x} \cdot \frac {\sin x}{\cos x(\sin x \cos x-1)}$

$\displaystyle \frac{\sin^2 x}{\cos^2 x}$

$\displaystyle \tan^2 x$
• Nov 8th 2008, 01:01 PM
Raj
Thanks masters

Quote:

Originally Posted by flyingsquirrel
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 $\displaystyle \sin^2x-\tan x$.

I was told I can't balance the identity (multiply/divide both sides) because I'm not initially told if it is equal, I have to prove if it is or if it is not.