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Prove the Following Trigonometric Identities. (1/(Sec^2 Theta - Cos Theta) + 1/(Cosec^2 Theta - Sin^2 Theta)) Sin^2 Theta Cos^2 Theta = (1 - Sin^2 Theta Cos^2 Theta)/(2 + Sin^2 Theta + Cos^2 Theta) - Mathematics

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Question

Prove the following trigonometric identities.

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

Sum
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Solution

In the given question, we need to prove

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

Now using `sec theta = 1/ cos theta` and `cosec theta = 1/sin theta` in LHS we get

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

`= (1/((1 - cos^4 theta)/cos^2 theta) + 1/((1 - sin^4 theta)/sin^2 theta)) sin^2 theta cos^2 theta`

`= ((cos^2 theta)/(1 - cos^4 theta) + sin^2 theta/(1 - sin^4 theta)) sin^2 theta cos^2 theta`

Further using the identity `a^2 - b^2 = (a + b)(a- b)` we get

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

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

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

`= ((cos^4 theta(1 + sin^2 theta) + sin^4 theta(1 + cos^2 theta))/((1 + cos^2 theta)(1 + sin^2 theta)))`

Further using the identity `sin^2 theta + cos^2 theta = 1` we get

LHS = `((cos^4 theta + cos^4 theta sin^2 theta + sin^4 theta + sin^4 theta cos^2 theta)/(1 + cos^2 theta + sin^2 theta + sin^2 theta cos^2 theta))`

`= (cos^4 theta + sin^4 theta + cos^2 theta sin^2 theta (cos^2 theta + sin^2 theta)) /(2 + sin^2 theta cos^2theta)`

`= ((cos^4 theta +sin^4 theta +cos^2 theta sin^2theta (1))/(2 + sin^2 theta cos^2 theta))`

Now, from the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get

So,

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

`= (((1)^2 - cos^2 theta sin^2 theta)/(22 +sin^2 theta cos^2 theta))`

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

Hence proved.

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Chapter 11: Trigonometric Identities - Exercise 11.1 [Page 45]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.1 | Q 57 | Page 45

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