Question

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

Answer 1

We have,

`LHS = q(p^2 – 1) = (secθ + cosecθ) [(sinθ + cosθ)^2 – 1]`

`=( \frac{1}{\cos \theta }+\frac{1}{\sin \theta } \right)\{\sin2\theta +\text{cos}2\theta +2\sin \theta \cos \theta 1\}`

`=( \frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta })(1+2\sin \theta \cos \theta 1)`

`=( \frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta })2\sin \theta \cos`

= 2(sinθ + cosθ) = 2p = RHS