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प्रश्न
sec6x - tan6x = 1 + 3sec2x × tan2x
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उत्तर
डावी बाजू = sec6x - tan6x
= (sec2x)3 - tan6x
= (1 + tan2x)3 - tan6x ......[∵ 1 + tan2θ = sec2θ]
= 1 + 3tan2x + 3(tan2x)2 + (tan2x)3 - tan6x .....[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
= 1 + 3tan2x (1 + tan2x) + tan6x - tan6x
= 1 + 3tan2x sec2x ......[∵ 1 + tan2θ = sec2θ]
= उजवी बाजू
∴ sec6x - tan6x = 1 + 3sec2x × tan2x
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संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
`1/(secθ - tanθ)` = secθ + tanθ
sec4θ - cos4θ = 1 - 2cos2θ
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
जर 3 sin θ = 4 cos θ, तर sec θ = ?
`costheta/(1 + sintheta) = (1 - sintheta)/(costheta)` हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
