Trig Identity Proof

**Coukapecker** · August 30th 2011, 03:08 PM

Hi,

I've been playing around with this thing for hours trying to get it to work and have made no progress ;\. anyone know how to do this?

Prove that for all angles θ,

tan(θ)^2/sin(θ)^2=1+tan(θ)^2

any help would be much appreciated,

thanks,

Coukapecker

**Siron** · August 30th 2011, 03:20 PM

Write (I use $x$ in stead of $\theta$ for a moment):
$\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}$
And also write:
$1+\tan^2(x)=1+\frac{\sin^2(x)}{\cos^2(x)}$ $=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=$ $\frac{1}{\cos^2(x)}$

Try to continue with this.

**Coukapecker** · August 30th 2011, 03:39 PM

That's actually where I got to without assistance, but I can't figure out what to do after

$tan^2(x)/sin^2(x)=sec^2(x)$

**Siron** · August 30th 2011, 03:42 PM

You have everything to prove the identity:
LHS:
$\frac{\frac{\sin^2(x)}{\cos^2(x)}}{\sin^2(x)}=$ $\frac{\sin^2(x)}{\cos^2(x)}\cdot \frac{1}{\sin^2(x)}=\frac{1}{\cos^2(x)}$

RHS:
$1+\tan^2(x)=\frac{1}{\cos^2(x)}$

So LHS=RHS

**Coukapecker** · August 30th 2011, 03:46 PM

Man,

I've bee staring at $(sin(x))^2/(cos(x))^2/(sin(x))^2)$ for like hours. Somehow managed to convince myself that you couldn't work with it.

Thanks for that :P.

**Siron** · August 30th 2011, 03:51 PM

You're welcome! Also in trigonometry you still have to know the important algebraic rules

.

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