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`

`1/cosθ+tanθ-sinθ1/cosθ-sinθtanθ=1/cosθ+tanθ-sinθ/cosθ-sinθtanθ`
We know that `tanθ=sinθ/cosθ` so let's replace all of the tanθ

`1/cosθ+sinθ/cosθ-sinθ/cosθ-sinθsinθ/cosθ`

`=1/cosθ+sinθ/cosθ-sinθ/cosθ-(sin^2θ)/cosθ`
the `+sinθ/cosθ-sinθ/cosθ=0` and we have:
`1/cosθ+0-sinθsinθ/cosθ=1/cosθ-(sin^2θ)/cosθ`


`=1/cosθ-(sin^2θ)/cosθ`
now we have common denominator of `cosθ` so we can write it as:
`(1-sin^2θ)/cosθ`
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:
`(cos^2θ)/cosθ=cosθ/1`

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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