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# Prove the Following Trigonometric Identities. ((1 + Sin Theta - Cos Theta)/(1 + Sin Theta + Cos Theta))^2 = (1 - Cos Theta)/(1 + Cos Theta) - CBSE Class 10 - Mathematics

#### Question

Prove the following trigonometric identities.

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

#### Solution

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.

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Solution Prove the Following Trigonometric Identities. ((1 + Sin Theta - Cos Theta)/(1 + Sin Theta + Cos Theta))^2 = (1 - Cos Theta)/(1 + Cos Theta) Concept: Trigonometric Identities.
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