Question

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

Answer 1

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.