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Question
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
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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
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sec4θ - cos4θ = 1 - 2cos2θ
secθ + tanθ = `cosθ/(1 - sinθ)`
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
sec2θ + cosec2θ = sec2θ × cosec2θ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sin2θ + sin2(90 – θ) = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
जर cos θ = `24/25`, तर sin θ = ?
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
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`
