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Question
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
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Solution
L.H.S. = sec6x – tan6x
L.H.S. = (sec2x)3 – tan6x
L.H.S. = (1 + tan2x)3 – tan6x ...[∵ 1 + tan2θ = sec2θ]
L.H.S. = 1 + 3tan2x + 3(tan2x)2 + (tan2x)3 – tan6x ...[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
L.H.S. = 1 + 3tan2x(1 + tan2x) + tan6x – tan6x
L.H.S. = 1 + 3tan2x.sec2x ...[∵ 1 + tan2θ = sec2θ]
R.H.S. = 1 + 3sec2x.tan2x
L.H.S. = R.H.S.
∴ sec6x – tan6x = 1 + 3sec2x × tan2x
Hence proved.
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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
