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प्रश्न
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
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उत्तर
डावी बाजू = `(tan^3θ - 1)/(tanθ - 1) = (tan^3θ - 1^3)/(tanθ - 1)`
= `((tanθ - 1)(tan^2θ + tanθ + 1))/((tanθ - 1))` ......…[∵ a3 – b3 = (a - b) (a2 + ab + b2)]
= tan2θ + tan θ + 1
= (1 + tan2θ) + tan θ
= sec2θ + tan θ ......…[∵ 1 + tan2θ = sec2θ]
= उजवी बाजू
∴ `(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
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संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
sec4A(1 - sin4A) - 2tan2A = 1
(sec θ + tan θ) (1 - sin θ) = cos θ
sec6x - tan6x = 1 + 3sec2x × tan2x
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
