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प्रश्न
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
उत्तर
We need to prove cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Solving the L.H.S, we get
`cosec^6 theta = (cosec^2 theta)^3`
`= (1 + cot^2 theta)^3` .......`(1 + cot^2 theta = cosec^2 theta)`
Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2` we get
`(1 + cot^2 theta)^3 = 1 + cot^6 theta + 3(1)^2 (cot^2 theta) + 3(1) (cot^2 theta)^2`
`= 1 + cot^6 theta + 3 cot^2 theta + 3 cot^4 theta`
`= 1 + cot^6 theta + 3 cot^2 theta (1 + cot^2 theta)`
`= 1 + cot^6 theta + 3 cot^2 theta cosec^2 theta` `(using 1 + cot^2 theta = cosec^2 theta)`
Hence proved.
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