# Proving Trigonometric Identities

• January 20th 2010, 02:10 PM
Quacky
Proving Trigonometric Identities
This question seems really easy, but I can't seem to work it out.

Prove that:
EDITTED: SEE BELOW

These are the identities we can use:
$Cos^2\theta+Sin^2\theta=1$
$\frac{Sin\theta}{cos\theta}=tan\theta$

$1+tan^2\theta=sec^2\theta$
$cosec^2\theta=1+cot^2\theta$

Thanks in advance for any help, please check back after posting as I may have a few problems.

Apologies, I've worked it out by myself. I'll edit the post if I have any other issues. Sorry for wasting anyone's time.
• January 20th 2010, 02:24 PM
$1+tan^2\theta=1+\frac{sin^2\theta}{cos^2\theta}=\f rac{cos^2\theta}{cos^2\theta}+\frac{sin^2\theta}{c os^2\theta}=\frac{sin^2\theta+cos^2\theta}{cos^2\t heta}$

$1+cot^2\theta=\frac{sin^2\theta}{sin^2\theta}+\fra c{cos^2\theta}{sin^2\theta}$
• January 20th 2010, 02:26 PM
Quacky
Sorry Archie (Doh), Thanks for confirming my answer at least.
• January 20th 2010, 02:36 PM
Soroban
Hello, Quacky!

Quote:

Prove that: . $\cot^2\theta + \cos^2\theta \;=\;(\csc \theta - \sin\theta)(\csc\theta + \sin\theta)$
I started with the right side . . .

$(\csc\theta -\sin\theta)(\csc\theta + \sin\theta) \;\;=\;\;\csc^2\!\theta - \sin^2\theta \;\;=\;\;\frac{1}{\sin^2\!\theta} - \sin^2\theta$

. . $=\;\; \frac{1-\sin^4\!\theta}{\sin^2\!\theta} \;\;=\;\;\frac{(1-\sin^2\!\theta)(1 + \sin^2\!\theta)}{\sin^2\!\theta} \;\;=\;\;\frac{\cos^2\!\theta(1 + \sin^2\!\theta)}{\sin^2\!\theta}$

. . $=\;\; \frac{\cos^2\!\theta + \sin^2\!\theta\cos^2\!\theta}{\sin^2\!\theta} \;\;=\;\;\frac{\cos^2\!\theta}{\sin^2\!\theta} + \frac{\sin^2\!\theta\cos^2\!\theta}{\sin^2\!\theta }$

. . $=\;\; \left(\frac{\cos\theta}{\sin\theta}\right)^2 + \cos^2\!\theta \;\;=\;\;\cot^2\!\theta + \cos^2\!\theta$

• January 20th 2010, 02:40 PM
Quacky
I feel really bad now (Crying), thanks though.
• January 20th 2010, 02:58 PM