Question

Simplify leaving your answer in terms of sin0(titha)

1 /1-cos0(titha) + 1 /1+cos0(titha)?

Answer 1

If your expression mean:

1 / (1 - cos θ ) + 1 / ( 1 + cos θ )

then

1 / (1 - cos θ ) + 1 / ( 1 + cos (θ) =

[ 1 ∙ ( 1 + cos θ ) + 1 ∙ ( 1 - cos θ ) ] / [ (1 - cos θ ) ∙ ( 1 + cos θ ) ] =

( 1 + cos θ + 1 - cos θ ) / ( 1 - cos² θ ) =

2 / sin² θ

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Remark:

( a - b ) ∙ ( a + b ) = a ² - b²

so

[ (1 - cos θ ) ∙ ( 1 + cos θ ) ] = 1² - cos² θ = 1 - cos² θ = sin² θ
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