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Questions
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following:
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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Solution
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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Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
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