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    I was wondering if anyone has any ideas how to answer this question?

    Any help would be great

    . Use the given triangle ( image of right angled triangle with sides labelled a,b,c and one angle labelled theta)

    to prove that for 0<(theta) <90
    1+tan squared theta = 1 over cos squared theta
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    It's probably easier to prove that \sin^2 \theta + \cos^2 \equiv 1 first, and then just divide through by \cos^2 \theta. Say that the hypotenuse has length 1, and then use SOH-CAH-TOA (to find the lengths of the other two sides in terms of \theta) and then Pythagoras's theorem (to get the required equation).
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    Use pythagoras to show sin^2 (theta) + cos^2 (theta)=1, then divide through by cos^2 theta to get the next equation.
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    (Original post by KellyLouiseee)
    I was wondering if anyone has any ideas how to answer this question?

    Any help would be great

    . Use the given triangle ( image of right angled triangle with sides labelled a,b,c and one angle labelled theta)

    to prove that for 0<(theta) <90
    1+tan squared theta = 1 over cos squared theta
    not sure if this is cheating but im assuming you still get taught the basic trig identity that sin^2x + cos^2x = 1
    divide both sides by cos^x

    you'll get 1 + tan^2x = 1/cos^2x (known as sec^2x)
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    c^2 = a^2 + b^2

    and tan theta = b/a and cos theta = c/a That is in my head c - hypo, b - opp, and a - adj

    See if that nudges you in the right direction.
 
 
 
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