Question

Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2

Answer 1

LHS = (sinθ + 1 + cosθ)(sinθ − 1 + cosθ). secθcosecθ

= [sin²θ − sinθ + sinθcosθ + sinθ − 1 + cosθ + sinθcosθ − cosθ + cos²θ] `1/cosθ1/sinθ `                                                             ...(∵ secθ = `1/cosθ  and  cosecθ = 1/sinθ`)

= [1 + 2sinθcosθ − 1]`1/cosθ  1/sinθ`

= [2sinθcosθ]`1/cosθ1/sinθ`

= 2 = RHS             

Hence proved.