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Question
tan4θ + tan2θ = sec4θ - sec2θ
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Solution
डावी बाजू = tan4θ + tan2θ
= `tan^2θ(tan^2θ + 1)`
= tan2θ.sec2θ ....[∵ 1 + tan2θ = sec2θ]
= `(sec^2θ - 1)sec^2θ` .....[∵ `tan^2θ = sec^2θ - 1`]
= sec4θ - sec2θ
= उजवी बाजू
∴ tan4θ + tan2θ = sec4θ - sec2θ
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RELATED QUESTIONS
cot θ + tan θ = cosec θ sec θ
sec4θ - cos4θ = 1 - 2cos2θ
sec2θ + cosec2θ = sec2θ × cosec2θ
cot2θ - tan2θ = cosec2θ - sec2θ
जर 1 – cos2θ = `1/4`, तर θ = ?
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= उजवी बाजू
`sec"A"/(tan "A" + cot "A")` = sin A 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
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
