SOLUTION: (sin^2θ -tanθ)/(cos^2θ-cotθ) = tan^2θ

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Question 732084: (sin^2θ -tanθ)/(cos^2θ-cotθ) = tan^2θ
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28sin%5E2%28theta%29-tan%28theta%29%29%2F%28cos%5E2%28theta%29-cot%28theta%29%29 = tanē(q)

In the left side:

Replace tan(q) with sin%28theta%29%2Fcos%28theta%29
and
Replace cot(q) with cos%28theta%29%2Fsin%28theta%29



Write as a division:

%28sin%5E2%28theta%29-sin%28theta%29%2Fcos%28theta%29%29%22%F7%22%28cos%5E2%28theta%29-cos%28theta%29%2Fsin%28theta%29%29

Get a common denominator in each:

%28sin%5E2%28theta%29cos%28theta%29%2Fcos%28theta%29-sin%28theta%29%2Fcos%28theta%29%29%22%F7%22%28cos%5E2%28theta%29sin%28theta%29%2Fsin%28theta%29-cos%28theta%29%2Fsin%28theta%29%29

%28sin%5E2%28theta%29cos%28theta%29-sin%28theta%29%29%2Fcos%28theta%29%22%F7%22%28cos%5E2%28theta%29sin%28theta%29-cos%28theta%29%29%2Fsin%28theta%29

Invert the second fraction and change division to multiplication:

%28sin%5E2%28theta%29cos%28theta%29-sin%28theta%29%29%2Fcos%28theta%29%22%22%2A%22%22sin%28theta%29%2F%28cos%5E2%28theta%29sin%28theta%29-cos%28theta%29%29

Factor sin(q)out of left numerator
and factor cos(q) out of right denominator:

sin%28theta%29%28sin%28theta%29cos%28theta%29-1%29%2Fcos%28theta%29%22%22%2A%22%22sin%28theta%29%2F%0D%0A%28cos%28theta%29%28cos%28theta%29sin%28theta%29-1%29%29

The two expressions in parentheses will cancel:

sin%28theta%29%28cross%28sin%28theta%29cos%28theta%29-1%29%29%2Fcos%28theta%29%22%22%2A%22%22sin%28theta%29%2F%0D%0A%28cos%28theta%29%28cross%28cos%28theta%29sin%28theta%29-1%29%29%29

%28sin%28theta%29%2Asin%28theta%29%29%2F%28cos%28theta%29%2Acos%28theta%29%29

sin%5E2%28theta%29%2Fcos%5E2%28theta%29

%28sin%28theta%29%2Fcos%28theta%29%29%5E2

tanē(q)

Edwin