prove `(1-sinθ)(secθ + tanθ) = 1/sec`

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prove `(1-sinθ)(secθ + tanθ) = 1/sec`

Postby barnamah » Thu 14 April, 2011 12:03

prove `(1-sinθ)(secθ + tanθ) = 1/sec`
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Re: prove `(1-sinθ)(secθ + tanθ) = 1/sec`

Postby Galaxy » Thu 14 April, 2011 12:22

we have `(1-sinθ)(secθ + tanθ) = 1/sec` meaning that we want `(1-sinθ)(secθ + tanθ)` is the same as `1/sec`
we have:`(1-sinθ)(secθ + tanθ)`
FOIL it and we have:
`(1-sinθ)(secθ + tanθ)=secθ+tanθ-sinθsecθ-sinθtanθ` we know `secθ=1/cosθ` so convert both secθ to `1/cosθ`

We know that `tanθ=sinθ/cosθ` so let's replace all of the tanθ


the `+sinθ/cosθ-sinθ/cosθ=0` and we have:

now we have common denominator of `cosθ` so we can write it as:
from trig identity we know `sin^2θ+cos^2θ=1` so `cos2θ=1-sin^2θ` therefore we can use this in our equation and we got:

Because `1/cosθ=secθ/1` so `1/secθ=cosθ` and that is what we got `cosθ/1` or cosθ
so `cosθ=1/secθ`
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