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प्रश्न
Prove that `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2 = 7 + tan^2 θ + cot^2 θ`.
बेरीज
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उत्तर
L.H.S = `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2`
=`(sin^2 θ + cosec^2 θ + 2 sin θ cosec θ + cos^2 θ + sec^2 θ + 2 cos θ sec θ)`
=`(sin^2 θ +cos^2 θ) + (cosec^2 θ + sec^2 θ) + 2sin θ (1/sin θ) + 2 cos θ (1/cos θ)`
= `(1) + (1 + cot^2 θ + 1 + tan^2 θ) + (2) + (2)`
= `7 + tan^2 θ + cot^2 θ`
= R .H. S
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