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Thread: Proving trig relationship

  1. #1
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    Question 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
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    Re: Proving trig relationship

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

    start with the Pythagorean identity ...

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

    ... and divide every term by $\cos^2{\theta}$
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    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.
    Last edited by otownsend; May 28th 2017 at 03:24 PM.
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    Re: Proving trig relationship

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

    $1+\tan^2{t} = \dfrac{\cos^2{t}}{\cos^2{t}}+ \dfrac{\sin^2{t}}{\cos^2{t}}=\dfrac{\cos^2{t}+\sin ^2{t}}{\cos^2{t}}= \dfrac{1}{\cos^2{t}}=\sec^2{t}$
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    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) ?
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    Re: Proving trig relationship

    Quote Originally Posted by otownsend View Post
    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.
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