Trigonometric Identities

**Ashz** · April 1st 2012, 04:12 PM

(sin²θ-cos²θ)/(1-tan²θ)=-cos²θ

So, the above is the equation that I have to verify.
I think the first step to this equation is

-1 (sin²θ-cos²θ)/(1-tan²θ)=-cos²θ
-1 (-(1))/(-1(sec²θ)=-cos²θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong (

)

1/(-1(sec²θ) = -cos²θ
Then by simplifying the above you get -cos²θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

Please help

**Soroban** · April 1st 2012, 04:47 PM

Hello, Ashz!

$\text{Verify: }\:\frac{\sin^2\!\theta - \cos^2\!\theta}{1 - \tan^2\!\theta} \:=\:-\cos^2\!\theta$

We have: . $\frac{\sin^2\!\theta - \cos^2\!\theta}{1 - \frac{\sin^2\!\theta}{\cos^2\!\theta}}$

$\text{Multiply by }\tfrac {\cos^2\theta}{\cos^2\theta}: \;\;\frac{\cos^2\!\theta\;\!(\sin^2\!\theta - \cos^2\!\theta)}{\cos^2\!\theta\left(1 - \frac{\sin^2\!\theta}{\cos^2\!\theta}\right)} \;=\;\frac{\cos^2\!\theta\;\!(\sin^2\!\theta - \cos^2\!\theta)}{\cos^2\!\theta - \sin^2\!\theta}$

. . . . . . . . . . . $=\;\frac{-\cos^2\!\theta\;\!(\cos^2\!\theta - \sin^2\!\theta)}{\cos^2\!\theta - \sin^2\!\theta} \;=\; -\cos^2\!\theta$

**Sudharaka** · April 1st 2012, 04:50 PM

Originally Posted by Ashz

(sin²θ-cos²θ)/(1-tan²θ)=-cos²θ

So, the above is the equation that I have to verify.
I think the first step to this equation is

-1 (sin²θ-cos²θ)/(1-tan²θ)=-cos²θ

This is incorrect. You have only multiplied the left hand side by -1.

-1 (-(1))/(-1(sec²θ)=-cos²θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong (

)

You have been given $(sin^{2}\theta-cos^{2}\theta)/(1-tan^{2}\theta)=-cos^{2}\theta$ If you want to multiply by -1 you have to multiply both the left had side and the right hand side.

1/(-1(sec²θ) = -cos²θ
Then by simplifying the above you get -cos²θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

Please help

Hi Ashz,

Consider the left hand side of this equation. Substitute, $\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2} \theta}$ and simplify.

**Prove It** · April 1st 2012, 04:51 PM

Originally Posted by Ashz

(sin²θ-cos²θ)/(1-tan²θ)=-cos²θ

So, the above is the equation that I have to verify.
I think the first step to this equation is

-1 (sin²θ-cos²θ)/(1-tan²θ)=-cos²θ
-1 (-(1))/(-1(sec²θ)=-cos²θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong (

)

1/(-1(sec²θ) = -cos²θ
Then by simplifying the above you get -cos²θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

Please help

$\displaystyle \begin{align*} \frac{\sin^2{x} - \cos^2{x}}{1 - \tan^2{x}} &\equiv \frac{\sin^2{x} - \cos^2{x}}{1 - \frac{\sin^2{x}}{\cos^2{x}}} \\ &\equiv \frac{\sin^2{x} - \cos^2{x}}{\frac{\cos^2{x} - \sin^2{x}}{\cos^2{x}}} \\ &\equiv \left(\sin^2{x} - \cos^2{x}\right) \left( \frac{\cos^2{x}}{\cos^2{x} - \sin^2{x}} \right) \\ &\equiv -\left(\cos^2{x} - \sin^2{x}\right)\left(\frac{\cos^2{x}}{\cos^2{x} - \sin^2{x}}\right) \\ &\equiv -\cos^2{x} \end{align*}$

Q.E.D.

**Ashz** · April 1st 2012, 05:03 PM

ahhh ~ that makes so much sense now!!! Thank you!!!!

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