Asked by Isha
sinθ / cosθ+sinθ - cosθ/ cosθ - sinθ
=tan²θ+1 / tan²θ -1
=tan²θ+1 / tan²θ -1
Answers
Answered by
oobleck
ever hear of parentheses? I assume you want to show that
sinθ/(cosθ+sinθ) - cosθ/(cosθ - sinθ) = (tan²θ+1) / (tan²θ -1)
using a common denominator, the left side is
[sinθ(cosθ-sinθ) - cosθ(cosθ+sinθ)]/(cos²θ-sin²θ)
= -(sin²θ+cos²θ)/(cos²θ-sin²θ)
= 1/(sin²θ-cos²θ)
Now divide top and bottom by cos²θ
and remember that sec²θ = 1+tan²θ
sinθ/(cosθ+sinθ) - cosθ/(cosθ - sinθ) = (tan²θ+1) / (tan²θ -1)
using a common denominator, the left side is
[sinθ(cosθ-sinθ) - cosθ(cosθ+sinθ)]/(cos²θ-sin²θ)
= -(sin²θ+cos²θ)/(cos²θ-sin²θ)
= 1/(sin²θ-cos²θ)
Now divide top and bottom by cos²θ
and remember that sec²θ = 1+tan²θ
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