Question

can someone explain this to me???

Answer 1

simplify

Answer 2

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?

Answer 3

\[1+\cos \theta -\cos \theta + \cos^2 ?\]

Answer 4

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

Answer 5

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

Answer 6

yes

Answer 7

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?

Answer 8

yes. but cant i break that down further?

Answer 9

are you understand @kapukawai

Answer 10

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:
\[\large \sqrt{\frac{\sin^{2}\theta}{\cos^{2}\theta}}\]

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

thank you!

Answer 15

You're welcome :)