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प्रश्न
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
उत्तर
We need to prove `sec^6 theta = tan^6 theta + 3 tan^2 theta sec^2 theta + 1`
Solving the L.H.S, we get
`sec^6 theta = (sec^2 theta)^3`
`= (1 + tan^2 theta)^3`
Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2` , we get
`(1 + tan^2 theta)^3 = 1 + tan^6 theta + 3(1)^2 (tan^2 theta) + 3(1)(tan^2 theta)^2`
`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`
`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`
`= 1 + tan^6 theta + 3 tan^2 theta (1 + tan^2 theta)`
`= 1 + tan^6 theta + 3 tan^2 theta sec^2 theta` (using `1 + tan^2 theta = sec^2 theta`)
Hence proved.
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