## anonymous one year ago can someone explain this to me???

1. anonymous

simplify

2. kropot72

The first step in simplifying is to multiply out the numerator: $\large (1-\cos \theta)(1+\cos \theta)=?$ Can you do that and post the result?

3. anonymous

$1+\cos \theta -\cos \theta + \cos^2 ?$

4. DecentNabeel

(1−cosθ)(1+cosθ)=1-cos^2x $\sqrt{\frac{1-\cos^2x}{\cos^2x}}$ $\mathrm{Use\:the\:following\:identity}:\quad \:1-\cos ^2\left(x\right)=\sin ^2\left(x\right)$ $=\sqrt{\frac{\sin ^2\left(x\right)}{\cos ^2\left(x\right)}}$

5. anonymous

ok. so im left with sin(x) over cos(x)?

6. DecentNabeel

yes

7. kropot72

Nearly correct! However the sign in front of the cos^2 (theta) term should be negative. Therefore we get: $\large \sqrt{\frac{(1-\cos^{2} \theta)}{\cos^{2} \theta}}$ Does that make sense?

8. anonymous

yes. but cant i break that down further?

9. DecentNabeel

are you understand @kapukawai

10. kropot72

You need to use the relationship: $\large \sin^{2}\theta+\cos^{2}\theta=1$ From which you get: $\large 1-\cos^{2}\theta=\sin^{2}\theta$ and finally by substitution we can write: $\large \sqrt{\frac{\sin^{2}\theta}{\cos^{2}\theta}}$

11. kropot72

Next you can use the relationship $\large \frac{\sin\theta}{\cos\theta}=\tan\theta$

12. anonymous

so tan theta is the simplified form of ...... ok that makes sence. i just have to memorise the relation ships?

13. kropot72

Yes, those relationships are some of the basic ones in trigonometry.

14. anonymous

thank you!

15. kropot72

You're welcome :)