Question

prove that:1-sinthetha/1+sinthetha=(sec^2thetha-tan^2thetha)​

Answer 1

solution :

To prove:

1-sin/1+sin = (sec- tan)^2

proof:

Taking RHS

(sec- tan)^2 => {1/cos - Sin/cos}^2

=> [ (1 - sin)/cos ]^2

=> (1 - Sin )^2 /cos^2

We know that sin square theta + cos square theta equal to 1 that implies cos square theta equal to 1 minus sin square theta

(1-sin)^2 / 1 - sin^2

cancelling 1 - sin'2 theta (a2-b2 = (a+b)(a-b)

=> > 1 - sin/1+sin = LHS