# Math Help - Trigonometric Identity Equation

1. ## Trigonometric Identity Equation [Solved]

Hello!

Firstly, thank you for looking at my question! I am new here and although I seem to be able to pick up mathematical concepts rather well, this one problem is absolutely impossible for me to figure out. I know the answer - I just don't know how to get there! I created an image of my problem so its easier to see what I am talking about. Here is the equation:

Essentially, what I am able to get is this:

(sin^2 x)/(cos^2 x) = tan^2 x
-(tan x)/ -(cot x) = 1

tan^2 x - 1 does not equal tan^2 x

Any help would be super greatly appreciated, thank you very much!
-Joshua

2. ## working with fractions

At first glance it looks like you are doing the following:
$
(6-1)/(3-1) = (6/3) - (1/1)
$

which isn't valid. Try again, paying attention to the algebra, not the trig.

This isn't relevant to your problem, but tan / cot doesn't equal 1.

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

$sin^2x-tanx = \frac{sin^2x}{cos^2x}(cos^2x - \frac{1}{tanx})$

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

$sin^2x-tanx = sin^2x-tanx$

4. Hello, vulcangon!

Welcome abaord!

$\frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \:=\:\tan^2\!x$

$\frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \;=\;\frac{\sin^2\!x - \dfrac{\sin x}{\cos x}}{\cos^2\!x - \dfrac{\cos x}{\sin x}}$

. . . . . . . . $= \;\frac{\sin x\left(\sin x - \dfrac{1}{\cos x}\right)} {\cos x\left(\cos x - \dfrac{1}{\sin x}\right)}$

. . . . . . . . $=\; \frac{\sin x\left(\dfrac{\sin x\cos x - 1}{\cos x}\right)} {\cos x\left(\dfrac{\sin x\cos x - 1}{\sin x}\right)}$

. . . . . . . . $=\; \frac{\dfrac{\sin x}{\cos x}(\sin x\cos x - 1)}{\dfrac{\cos x}{\sin x}(\sin x\cos x - 1)}$

. . . . . . . . $= \;\frac{\tan x}{\cot x}$

. . . . . . . . $=\; \frac{\tan x}{\frac{1}{\tan x }}$

. . . . . . . . $=\;\tan^2\!x$

5. Originally Posted by Soroban
Hello, vulcangon!

Welcome abaord!

$\frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \;=\;\frac{\sin^2\!x - \dfrac{\sin x}{\cos x}}{\cos^2\!x - \dfrac{\cos x}{\sin x}}$

. . . . . . . . $= \;\frac{\sin x\left(\sin x - \dfrac{1}{\cos x}\right)} {\cos x\left(\cos x - \dfrac{1}{\sin x}\right)}$

. . . . . . . . $=\; \frac{\sin x\left(\dfrac{\sin x\cos x - 1}{\cos x}\right)} {\cos x\left(\dfrac{\sin x\cos x - 1}{\sin x}\right)}$

. . . . . . . . $=\; \frac{\dfrac{\sin x}{\cos x}(\sin x\cos x - 1)}{\dfrac{\cos x}{\sin x}(\sin x\cos x - 1)}$

. . . . . . . . $= \;\frac{\tan x}{\cot x}$

. . . . . . . . $=\; \frac{\tan x}{\frac{1}{\tan x }}$

. . . . . . . . $=\;\tan^2\!x$

Ahhhhhhh Thank you!

I get it now.

Also, thanks everyone who responded

6. But I don't get this so well

$3sin^4 x - 2sin^6 x = 1 - 3cos^4 x + 2cos^6 x$