Advertisements
Advertisements
Question
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
Solution
डावी बाजू = `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2`
= `tanA/(sec^2A)^2 + cotA/("cosec"^2A)^2` ...`[(∵ 1 + tan^2θ = sec^2θ","),(1 + cot^2θ = "cosec"^2θ)]`
= `tanA/(sec^4A) + cotA/("cosec"^4A)`
= `tanA xx 1/(sec^4A) + cotA xx 1/("cosec"^4A)`
= `sinA/cosA xx cos^4A + cosA/sinA xx sin^4A`
= sinA cos3A + cosA sin3A
= sinA cosA (cos2A + sin2A)
= sinA cosA (1) ...[∵ sin2θ + cos2θ = 1]
= sinA cosA
= उजवी बाजू
∴ `tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sinA cosA
APPEARS IN
RELATED QUESTIONS
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sec2θ – tan2θ = ?
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A 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
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
