Prove Trig Identity

**Raj** · November 8th 2008, 11:52 AM

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

I've tried and cannot prove it

**flyingsquirrel** · November 8th 2008, 12:26 PM

Hi,

This equality can be written $\sin^2x-\tan x =\tan^2x\left(\cos^2x-\cot x\right)$ . You should try to use $\begin{cases}\tan x = \frac{\sin x }{\cos x}\\ \cot x =\frac{1}{\tan x}\end{cases}$ to show that $\tan^2x\left(\cos^2x-\cot x\right)$ equals $\sin^2x-\tan x$ .

**masters** · November 8th 2008, 12:26 PM

Originally Posted by Raj

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

I've tried and cannot solve it

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

Working with Left Hand Side:

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

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

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

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

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

$\tan^2 x$

**Raj** · November 8th 2008, 01:01 PM

Thanks masters

Originally Posted by flyingsquirrel

Hi,

This equality can be written $\sin^2x-\tan x =\tan^2x\left(\cos^2x-\cot x\right)$ . You should try to use $\begin{cases}\tan x = \frac{\sin x }{\cos x}\\ \cot x =\frac{1}{\tan x}\end{cases}$ to show that $\tan^2x\left(\cos^2x-\cot x\right)$ equals $\sin^2x-\tan x$ .

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.

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