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If cosec θ + cot θ = p, then prove that cos θ = p2-1p2+1 - Mathematics

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प्रश्न

If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`

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उत्तर

According to the question,

cosec θ + cot θ = p

Since, cosec θ = `1/sintheta` and cot θ = `costheta/sintheta`

`1/sintheta + costheta/sintheta` = p

`(1 + costheta)/sintheta` = p

Squaring on L.H.S and R.H.S,

`((1 + costheta)/sin theta)^2` = p2

`(1 + cos^2 theta + 2 cos theta)/(sin^2 theta)` = p2

Applying component and dividend rule,

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

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

Since, 1 – sin2θ = cos2θ and sin2θ + cos2θ = 1

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

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

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

cos θ = `("p"^2 - 1)/("p"^2 + 1)`

Hence proved.

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पाठ 8: Introduction To Trigonometry and Its Applications - Exercise 8.4 [पृष्ठ ९९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
पाठ 8 Introduction To Trigonometry and Its Applications
Exercise 8.4 | Q 1 | पृष्ठ ९९

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∴ sin θ =  `9/41`

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