Q. 45

# If sinθ + cosθ = p and secθ + cosecθ = q, then show q(p2–1) =2p

Given: sin θ + cos θ = p and sec θ + cosec θ = q

To show q(p2 – 1) = 2p

Taking LHS = q(p2 – 1)

Putting the value of sin θ + cos θ = p and sec θ + cosec θ = q, we get

=(sec θ + cosec θ){( sin θ + cos θ )2 – 1)

=(sec θ + cosec θ){(sin2 θ + cos2 θ + 2sin θ cosθ) – 1)}

[ (a + b)2 = (a2 + b2 + 2ab)]

=(sec θ + cosec θ)(1+2sin θ cos θ – 1)

=(sec θ + cosec θ)(2sin θcosθ)

= 2(sin θ +cos θ)

= 2p [ given sin θ + cos θ = p]

=RHS

Hence Proved

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