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Prove the following: sec6x – tan6x = 1 + 3sec2x × tan2x

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प्रश्न

Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x

योग
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उत्तर

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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अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 5.07 | पृष्ठ १३८

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