Question

If sin θ + cos θ = m and sec θ + cosec θ = n, prove that n(m² – 1) = 2m

Answer 1

Given:

sin θ + cos θ = m and secθ + cosecθ = n

Consider L.H.S. = n(m² − 1) = (secθ + cosecθ)[(sinθ + cosθ)² − 1]

= `(1/cosθ + 1/sinθ) [sin^2θ + cos^2θ + 2sinθcosθ - 1`]

= `((cosθ + sinθ)/(sinθcosθ)) (1 + 2sinθcosθ - 1)`

= `((cosθ + sinθ))/(sinθcosθ) (2 sinθ cosθ)`

= 2(sinθ + cosθ)

= 2m = R.H.S.