How Do You Prove This Trigonometric Identity?

[preet]

Hi... I need help solving this problem. I don't know what to do...

Proove that

\frac{cos^2\theta - sin^2\theta}{cos^2\theta+sin\theta cos\theta} = 1 - tan\theta

I tried to cross out cos^2 on the top with the one at the bottom... also tried messing around with the values (cos^2x = 1 - sin^2x) etc... but nothing is working. I'm kind of lost and would appreciate any help... Thanks

 

[courtrigrad]

First if you are given \frac{cos^2\theta - sin^2\theta}{cos^2\theta+sin\theta cos\theta} = 1 - tan\theta factor the numerator (its a difference of two perfect squares). Then factor the denominator (can you see a common term?) Cancel like terms. Divide through and you will get your result.

Hint: a^2 - b^2 = (a-b)(a+b)

 

[Ryoukomaru]

Cross out with cos^2(x) also works.


\frac{cos^2(x)-sin^2(x)}{cos^2(x)}
\frac{cos^2(x)+sin(x)cox(x)}{cos^2(x)}

When you solve these, the answer comes out very nicely

 

[preet]

maybe it's because it's only been a little over a week from the xmas break, but even after factoring, I don't know the next step... how can i cancel out stuff still in the brackets? And for crossing it out... I don't really get what happened. What did you cross out exactly, and why is the fraction flipped?

factored...
\frac{(cos\theta + sin\theta) (cos\theta - sin\theta)}{(cos\theta+sin\theta)(cos\theta)}

 

[Ryoukomaru]

In the denominator, inside the brackets you get (cos\theta+sin\theta) because before factoring you had cos^2\theta. Now do some cancelling out and you will get the answer.

 

[preet]

yah, i just caught my mistake ... but I am still not getting 1-tan theta...
I end up with cos theta - sin theta over cos theta after cancelling...

 

[courtrigrad]

ok so you see \frac{(cos\theta + sin\theta) (cos\theta - sin\theta)}{(cos\theta+sin\theta)(cos\theta)} (cos\theta+sin\theta) both in the numerator and denominator. So cancel that out and you are left with:

\frac{(cos\theta - sin\theta)}{(cos\theta)}

Simplify this and you get the answer

 

[Ryoukomaru]

\frac{cos\theta-sin\theta}{cos\theta}=\frac{cos\theta}{cos\theta}-\frac{sin\theta}{cos\theta}

 

[Curious3141]

What is the definition of \tan \theta in terms of \sin \theta and \cos \theta ?

 

[preet]

mmmannnn... I think I need another vacation. Thanks for the help =)