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Question
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
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Solution
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
L.H.S = [(sin θ + sec θ)2 + (cos θ + cosec θ)2]
= [sin2 θ + sec2 θ + 2 sin θ sec θ + cos2 θ + cosec2 θ + 2 cos θ cosec θ]
= (sin2θ + cos2θ) + (sec2θ + cosec2θ) + 2 (sinθ secθ + cos θ cosec θ)
= `1 + sec^2 theta + "cosec"^2 theta + 2[sin theta xx 1/cos theta + cos theta xx 1/sin theta]`
= `1 + sec^2 theta + "cosec"^2 theta + 2 [(sin^2 theta + cos^2 theta)/(sintheta cos theta)]`
= `1 + sec^2 theta + "cosec"^2 theta + 2 xx 1/(sintheta costheta)`
= 1 + sec2θ + cosec2θ + 2 secθ cosecθ
= 1 + (secθ + cosecθ)2
L.H.S = R.H.S
∴ (sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
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