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Question
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
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Solution
L.H.S. `(1 - 2sin^2A)^2/(cos^4A - sin^4A)`
= `(1 - 2sin^2A)^2/((cos^2A)^2 - (sin^2A)^2`
= `(1 - 2sin^2A)^2/((cos^2A + sin^2A)(cos^2A - sin^2A))`
= `(1 - 2sin^2A)^2/((1)(1 - sin^2A - sin^2A)` ...[∵ cos2 A = 1 – sin2 A]
= `(1 - 2sin^2A)^2/((1 - 2sin^2A))`
= 1 – 2 sin2 A
= 1 – 2 (1 – cos2 A)
= 1 – 2 + 2 cos2 A
= 2 cos2 A – 1 = R.H.S.
Hence the result is proved.
RELATED QUESTIONS
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(secA + tanA) (1 − sinA) = ______.
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`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
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`(cosecA - 1)/(cosecA + 1) = (cosA/(1 + sinA))^2`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Define an identity.
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
