Question
[Sin^2(theta) - tan(theta) ] / [cos^2(theta) - cot(theta)] = tan^2(theta)
I am supposed to prove that the left side of the equation is supposed to be equal to the right side. And I been confused on how to solve it.
I am supposed to prove that the left side of the equation is supposed to be equal to the right side. And I been confused on how to solve it.
Answers
As a general rule, I change all trig ratios to sines or cosines unless I recognize
a definite identity, so ...
LS = (sin^2 θ - sinθ/cosθ) / (cos^2 θ - cosθ/sinθ)
= [ (cosθsin^2 θ - sinθ)/cosθ ] / [ (sinθcos^2 θ - cosθ)/sinθ ]
= (cosθsin^2 θ - sinθ)/cosθ * sinθ /(sinθcos^2 θ - cosθ)
= sinθ(sinθcosθ - 1)/cosθ * sinθ/(cosθ(sinθcosθ - 1) )
= sin^2 θ / cos^2 θ
= tan^2 θ
= RS
a definite identity, so ...
LS = (sin^2 θ - sinθ/cosθ) / (cos^2 θ - cosθ/sinθ)
= [ (cosθsin^2 θ - sinθ)/cosθ ] / [ (sinθcos^2 θ - cosθ)/sinθ ]
= (cosθsin^2 θ - sinθ)/cosθ * sinθ /(sinθcos^2 θ - cosθ)
= sinθ(sinθcosθ - 1)/cosθ * sinθ/(cosθ(sinθcosθ - 1) )
= sin^2 θ / cos^2 θ
= tan^2 θ
= RS
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