Trigonometry Pproblems Help!!!!!! *10 Pointsss**?

Use trigonometric identities to transform the left side of the equation to the right one.

*** I'm using Q as theta***

1) tanQ CosQ= SinQ

2) (1+CosQ)(1-CosQ)=sin^2Q

3) (SecQ+TanQ)(SecQ-TanQ)=1

4) SinQ/CosQ=CscQ SecQ

1 Answer

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  • 1 decade ago
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    1.

    Starting from the left hand side (LHS):

    = tanθ.cosθ

    = (sinθ/cosθ).cosθ

    Cancel the common cosθ from numerator and denominator:

    = sinθ

    = RHS...Proved

    2.

    Starting from the left hand side (LHS):

    = (1+cosθ)(1-cosθ)

    This is essentially (a+b)(a-b) = a²-b²

    = 1-cos²θ

    = sin²θ

    [sin²θ+cos²θ = 1, this is the Pythagorean identity. Just subtract cos²θ from both sides of this identity]

    = RHS...Proved

    3.

    Starting from the left hand side (LHS):

    = (secθ+tanθ)(secθ-tanθ)

    = sec²θ-tan²θ [same as in #2]

    = 1 [same as in #2, sec²θ-tan²θ = 1]

    = RHS...Proved

    Alternatively, convert sec and tan to their respective sine/cosine forms, and you will see that they are the same:

    = sec²θ-tan²θ

    = (1/cos²θ)-(sin²θ/cos²θ)

    = (1-sin²θ)/cos²θ

    = cos²θ/cos²θ

    = 1

    4.

    Not sure about this one. Is the question correct?

    sinθ/cosθ = tanθ and cscθ.secθ = 1/(sinθ.cosθ).

    Can't think of a way to prove 1/(sinθ.cosθ) = tanθ.

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