Question

Prove the following trigonometric identities.

`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`

Answer 1

 In the given question, we need to prove `((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`

Taking `sin theta` common from the numerator and the denominator of the L.H.S, we get

`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2  = (((sin theta)(cosec theta + 1 -cot theta))/((sin theta)(cosec theta + 1 + cot theta)))^2`

`= ((1 + cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`

Now, using the property  `1 + cot^2 theta = cosec^2 theta` we get

`((1 + cosec theta -  cot theta)/(1 + cosec theta + cot theta))^2 = (((cosec^2 theta - cot^2 theta) +cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`

Using `a^2 - b^2 = (a + b)(a - b) we get

`(((cosec^2 theta - cot^2 theta)(cosec theta - cot theta))/(1 + cosec theta + cot theta))^2 = (((cosec theta - cot theta)(cosec theta + cot theta + 1))/(1 + cosec theta + cot theta))^2`

`= (cosec theta - cot theta)^2`

Using `cot theta = cos theta/sin theta` and `cosec = 1/sin theta` we get

`(cosec theta - cot theta)^2 = (1/sin theta - cos theta/sin theta)^2`

`= ((1 - cos theta)/sin theta)^2`

Now, using the property `sin^2 theta + cos^2 theta = 1` we get

`(1 - cos theta)^2/sin^2 theta = (1 - cos theta)/(1 - cos^2 theta)`

`= (1 - cos theta)^2/((1 + cos theta)(1 - cos theta))`

`= (1 - cos theta)/(1 + cos theta)`

Hence proved.