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Question
Prove the following:
`(2cos 4x + 1)/(2cosx + 1)` = (2 cos x – 1) (2 cos 2x – 1)
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Solution
L.H.S. = `(2cos 4x + 1)/(2cosx + 1)`
= `(2[2cos^2(2x) - 1] + 1)/(2cos x + 1)` ...[∵ cos 2θ = 2 cos2 θ – 1]
= `(4cos^2 2x - 2 + 1)/(2cos x + 1)`
= `((2cos 2x)^2 - (1)^2)/(2cos x + 1)`
= `((2cos 2x + 1)(2cos 2x - 1))/(2cos x + 1)`
= `([2(2cos^2 x - 1) + 1](2cos 2x - 1))/(2cosx + 1)`
= `((4cos^2 x - 2 + 1)(2cos 2x - 1))/(2cos x + 1)`
= `([(2cos x)^2 - (1)^2](2cos 2x - 1))/(2cos x + 1)`
= `((2cos x + 1)(2cos x - 1)(2cos 2x - 1))/(2cos x + 1)`
= (2 cos x – 1) (2 cos 2x – 1)
= R.H.S.
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