Question

Simplify
(1- cos θ)(1+cos θ) /(1-sin θ)(1+sin θ)

Answer 1

You are given \[\frac{(1 - \cos(\theta))(1 + \cos(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))}\] and asked to simplify the expression right?

Answer 2

Correct

Answer 3

\[(1- \cos θ)(1+\cos θ) \over (1-\sin θ)(1+\sin θ)\]

Answer 4

And in this case can apply the difference of squares rule \((a - b)(a + b) = a^2 - b^2\) so that:

\(\dfrac{(1 - \cos(\theta))(1 + \cos(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))} = \dfrac{1 - \cos^2(\theta)}{1 - \sin^2(\theta)}\) Do you agree?

Answer 5

Yes I do

Answer 6

Now from Pythagorean Identity \(\sin^2\theta + \cos^2 \theta = 1\), we know that

\(\sin^2 \theta = 1 - \cos^2\theta\)

and

\(\cos^2\theta = 1 - \sin^2\theta\)

right?

Answer 7

Yeah

Answer 8

Therefore

\[\dfrac{1 - \cos^2(\theta)}{1 - \sin^2(\theta)} = \dfrac{\sin^2\theta}{\cos^2\theta}\] Correct?

Answer 9

Yeah thank you so much for explaining

Answer 10

We're not done yet. Because of algebraic rule \(\dfrac{a^2}{b^2} = \left(\dfrac{a}{b}\right)^2\)

We can write \(\dfrac{\sin^2\theta}{\cos^2\theta} = \left(\dfrac{\sin\theta}{\cos\theta}\right)^2\). Do you agree?

Answer 11

I do agree

Answer 12

And we know that \(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\) therefore

\(\left(\dfrac{\sin\theta}{\cos\theta}\right)^2 = (\tan\theta)^2\)

Answer 13

Makes sense right?

Answer 14

Yes it does

Answer 15

So in conclusion should we should end by stating \(\dfrac{(1 - \cos(\theta))(1 + \cos(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))} = \tan^2\theta\) ?

Answer 16

Well thank you Hero for helping. I understand how to do it now.

Answer 17

You were a big help

Answer 18

You're welcome