1. ## Proving trig relationship

Hi,

I hope someone can help. I was hoping that someone can provide a proof for why 1 + tan^2θ does not equal sec^2
θ. I know that I can have θ equal something, and then see if both sides end up being the same e.g. 1 + tan^2(pi/3) = sec^2(pi/3) .... 1 + (√3)^2 = (1/(1/2))^2 ... 4 = 4.

While this case above (when θ = pi/3) proves that the equation is right, I don't think this proves why this is right for all cases of
θ. Can someone help provide a more general proof for why the LS equals the RS?

Sincerely,
Olivia

2. ## Re: Proving trig relationship

$1+\tan^2{\theta}$ does equal $\sec^2{\theta}$

$\cos^2{\theta}+\sin^2{\theta}=1$

... and divide every term by $\cos^2{\theta}$

3. ## Re: Proving trig relationship

Your response is exactly what I was looking for. Thank you

I understand how dividing cos^2θ will result in 1+tan^2θ = sex^2θ, but I'm curious how does one realize that this step would accomplish this? Is it just a matter of seeing relationships between trig functions or is there a sort of trick involved in knowing what steps you need to do in order prove or disprove it? I hope what I'm saying makes sense.

4. ## Re: Proving trig relationship

No trick ... just a shortcut. Here's a more formal derivation ...

$1+\tan^2{t} = \dfrac{\cos^2{t}}{\cos^2{t}}+ \dfrac{\sin^2{t}}{\cos^2{t}}=\dfrac{\cos^2{t}+\sin ^2{t}}{\cos^2{t}}= \dfrac{1}{\cos^2{t}}=\sec^2{t}$

5. ## Re: Proving trig relationship

Okay thanks. However, I don't see how you made the jump from (cos^2t+sin^2t)/(cos^2t) to 1/(cos^2t) ?

6. ## Re: Proving trig relationship

Originally Posted by otownsend
Okay thanks. However, I don't see how you made the jump from (cos^2t+sin^2t)/(cos^2t) to 1/(cos^2t) ?
Look up the Pythagorean identity in your text or elsewhere online.