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प्रश्न
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
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उत्तर
डावी बाजू = `(sintheta + "cosec" theta)/sin theta`
= `sintheta/sintheta + ("cosec"theta)/sintheta`
= 1 + cosec θ × cosec θ ......`[∵ "cosec" theta = 1/sin theta]`
= 1 + cosec2θ
= 1 + 1 + cot2θ .......[∵ 1 + cot2θ = cosec2θ]
= 2 + cot2θ
= उजवी बाजू
∴ `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
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संबंधित प्रश्न
sec4θ - cos4θ = 1 - 2cos2θ
1 + tan2θ = किती?
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
tan2θ – sin2θ = tan2θ × sin2θ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 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`
