मराठी

If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.

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प्रश्न

If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.

बेरीज
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उत्तर

Given that,

sin θ + cos θ = p       ...(i)

and sec θ + cosec θ = q

`\implies 1/cos θ + 1/sin θ` = q       ...`[∵ sec θ = 1/cos θ and "cosec"  θ = 1/sinθ]`

`\implies (sin θ + cos θ)/(sin θ . cos θ)` = q

`\implies "p"/(sin θ . cos θ)` = q           ...[From equation (i)]

`\implies` sin θ. cos θ = `"p"/"q"`         ...(ii)

sin θ + cos θ = p

On squaring both sides, we get

(sin θ + cos θ)2 = p2

`\implies` (sin2 θ + cos2 θ) + 2 sin θ . cos θ = p2      ...[∵ (a + b)2 = a2 + 2ab + b2]

`\implies` 1 + 2sin θ . cos θ = p2      ...[∵ sin2 θ + cos2 θ = 1]

`\implies` `1 + 2 . "p"/"q"` = p2         ...[From equation (iii)]

`\implies` q + 2p = p2q

`\implies` 2p = p2q – q

`\implies` q(p2 – 1) = 2p

Hence proved.

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

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 10
पाठ 8 Introduction To Trigonometry and Its Applications
Exercise 8.4 | Q 10 | पृष्ठ ९९
सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 6 Trigonometry
Exercise 6.1 | Q 9. (i) | पृष्ठ २५०

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