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Prove that Sqrt((1 + Cos Theta)/(1 - Cos Theta)) + Sqrt((1 - Cos Theta)/(1 + Cos Theta)) = 2 Cosec Theta - Mathematics

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प्रश्न

Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`

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उत्तर

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

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

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

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

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

`= 2/sin theta = 2cosec theta`

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पाठ 11: Trigonometric Identities - Exercise 11.1 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 11 Trigonometric Identities
Exercise 11.1 | Q 83.3 | पृष्ठ ४७

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cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= `cosθ/sinθ + square/cosθ`

= `(square + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ `square`

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Show that, cotθ + tanθ = cosecθ × secθ

Solution :

L.H.S. = cotθ + tanθ

= `cosθ/sinθ + sinθ/cosθ`

= `(square + square)/(sinθ xx cosθ)`

= `1/(sinθ xx cosθ)` ............... `square`

= `1/sinθ xx 1/square`

= cosecθ × secθ

L.H.S. = R.H.S

∴ cotθ + tanθ = cosecθ × secθ


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