# Prove that `tan(a+b) /cot(a-b)=(sin^2a-sin^2b)/(cos^2a-sin^2b)`

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### 1 Answer

The identity `tan(a+b) /cot(a-b)=(sin^2a-sin^2b)/(cos^2a-sin^2b)` has to be proved.

`tan(a+b) /cot(a-b)`

=> `tan(a+b)*tan(a -b)`

=> `(sin(a+b)/cos(a+b))*(sin(a-b)/cos(a-b))`

=> `(sin(a+b)*sin(a-b))/(cos(a+b)*cos(a-b))`

=>`((1/2)*cos(a+b-a+b) - (1/2)*cos 2a)/((1/2)*cos(a +b-a+b) + (1/2)*cos(a +b +a-b))`

=> `(cos(2*b) - cos 2a)/(cos(2*b) + cos(2*a))`

=> `(1 - 2*sin^2b - 1 + 2*sin^2a)/(2*cos^2b - 1 + 1 - 2*sin^2b)`

=> `(sin^2a - sin^2b)/(cos^2b - sin^2b)`

**This proves that `tan(a+b) /cot(a-b)=(sin^2a-sin^2b)/(cos^2a-sin^2b)` **