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Question
Prove that (sinθ + cosecθ)2 + (cosθ + secθ)2 = 7 + tan2θ + cot2θ.
Solution 1
LHS = (sinθ + cosecθ)2 + (cosθ + secθ)2
= sin2θ + cosec2θ + 2sinθ . cosecθ + cos2θ + sec2θ + 2cosθ . secθ ...[∵ (a + b)2 = a2 + 2ab + b2]
= `(sin^2θ + cos^2θ) + (1 + cot^2θ) + 2sinθ 1/sinθ + (1 + tan^2θ) + 2cosθ1/(cosθ)` `[∵ cosec^2θ = 1 + cot^2θ, sec^2θ = 1 + tan^2θ,
cosecθ = 1/(sinθ) and secθ = 1/cosθ ]`
= 1 + 1 + cot2θ + 2 + 1 + tan2θ + 2 ...[∵ sin2θ + cos2θ = 1]
= 7 + tan2θ + cot2θ
= RHS, Hence proved.
Solution 2
LHS = (sinθ + cosecθ)2 + (cosθ + secθ)2
= sin2θ + cosec2θ + 2sinθ . cosecθ + cos2θ + sec2θ + 2cosθ . secθ
= (sin2θ + cos2θ) + cosec2θ + sec2θ + 2 + 2 ...(∵ sinθ · cosecθ = 1 and secθ · cosθ = 1]
= 1 + cosec2θ + sec2θ + 4 ...[∵ sin2θ+ cos2θ = 1]
= 5 + (1 + cot2θ) + (1 + tan2θ)
= 7 + cot2θ + tan2θ = R.H.S.
