Trig identities help?

Prove the identity:

(cosθ - sinθ) / (cosθ + sinθ) = sec2(θ) - tan2(θ)

thanks so much! show work please.

Update:

Will award best answer!!!

3 Answers

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  • 5 months ago

    (cos(theta) - sin(theta)) / (cos(theta) + sin(theta))

    = (cos(theta) - sin(theta))^2 / (cos(theta) + sin(theta))(cos(theta) - sin(theta))

    = (cos^2(theta) - 2 sin(theta)cos(theta) + sin^2(theta)) / (cos^2(theta) - sin^2(theta))

    = (1 - 2 sin(theta)cos(theta)) / cos(2theta)

    = (1 - sin(2theta)) / cos(2theta)

    = 1/cos(2theta) - sin(2theta)/cos(2theta)

    = sec(2theta) - tan(2theta).

    Word of advice: might wanna move each left parenthesis over and include the 2 in each pair of parentheses. Someone may mistake the right side as sec^2(theta) - tan^2(theta), which BTW is 1. 

    As for proving your given identity the key is to multiply numerator and denominator by cos(theta) - sin(theta).  That leads to a difference of squares in the denominator, cos^2(theta) - sin^2(theta), which is an equivalent for the cosine of a double angle, cos(2theta).  In the numerator you have a perfect square, cos^2(theta) - 2 sin(theta)cos(theta) + sin^2(theta).  There you'd make 2 substitutions: cos^2(theta) + sin^2(theta) is 1, and -2 sin(theta)cos(theta) is -sin(2theta). (1 - sin(2theta)) / cos(2theta) results; from there break this up into a difference of two expressions with the same denominator and finish.

  • Anonymous
    5 months ago

    I'm not sure you've written it down right, it looks false.  If the RHS has squared in it for 2 then it's fdlse 

    Sec to power 2 minus tan to the power 2 is 1

  • ted s
    Lv 7
    5 months ago

    just multiply top and bottom by ' cos Θ + sin Θ ' and it " falls out "

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