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Question
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
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Solution
L.H.S = sec4 θ (1 – sin4 θ) – 2 tan2 θ
= `1/cos^4 theta [1 - (sin^2 theta)^2]- 2 xx (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) (1 + sin^2 theta) (1 - sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) xx cos^2 theta (1 + sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta)/(cos^2 theta) - (2sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta - 2sin^2 theta)/(cos^2 theta)`
= `(1 - sin^2 theta)/(cos^2 theta)`
= `(cos^2 theta)/(cos^2 theta)`
L.H.S = R.H.S
∴ sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
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Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
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= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
