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प्रश्न
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
उत्तर
डावी बाजू = sin2A . tan A + cos2A . cot A + 2 sin A . cos A
= `sin^2"A"* (sin "A")/(cos "A") + cos^2"A"* (cos"A")/(sin"A") + 2sin"A" *cos"A"`
= `(sin^3"A")/"cosA" + (cos^3"A")/"sinA" + 2sin"A"*cos"A"`
= `(sin^4"A" + cos^4"A" + 2sin^2"A"cos^2"A")/(sin"A"cos"A")`
= `(sin^2"A" + cos^2"A")^2/(sin"A"cos"A")` .....[∵ a2 + b2 + 2ab = (a + b)2]
= `1^2/(sin"A"cos"A")` ......[∵ sin2A + cos2A = 1]
= `1/(sin"A"cos"A")`
= `(sin^2"A"+ cos^2"A")/(sin"A"cos"A")` ......[∵ 1 = sin2A + cos2A]
= `(sin^2"A")/(sin"A"cos"A") + (cos^2"A")/(sin"A"cos"A")`
= `"sinA"/"cosA" + "cosA"/"sinA"`
= tan A + cot A
= उजवी बाजू
∴ sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
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संबंधित प्रश्न
cos2θ(1 + tan2θ) = 1
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
tan4θ + tan2θ = sec4θ - sec2θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
खालीलपैकी चुकीचे सूत्र कोणते?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"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
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
sin2θ + cos2θ ची किंमत काढा.
उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`