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