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Prove the Following Trigonometric Identities. (Cos Theta - Sin Theta + 1)/(Cos Theta + Sin Theta - 1) = Cosec Theta + Cot Theta

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प्रश्न

Prove the following trigonometric identities.

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

प्रमेय
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उत्तर

We have to prove the following identity

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

Consider the LHS = `(cos theta - sin theta + 1)/(cos theta + sin theta - 1)`

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

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

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

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

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

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

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

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

`= cos theta/sin theta + 1/sin theta`

`= cot theta + cosec theta`

= RHS

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अध्याय 11: Trigonometric Identities - EXERCISE 11.1 [पृष्ठ ११.३५]

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आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 11 Trigonometric Identities
EXERCISE 11.1 | Q 35. (iii) | पृष्ठ ११.३५

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∴ 1 + tan2 θ = `(25/7)^square`

∴ tan2 θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)


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