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Question
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
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Solution
डावी बाजू = cot2θ – tan2θ
= (cosec2θ − 1) − (sec2θ − 1) ......`[(because tan^2theta = sec^2theta - 1),(cot^2theta = "cosec"^2 theta - 1)]`
= cosec2θ − 1 − sec2θ + 1
= cosec2θ − sec2θ
= उजवी बाजू
∴ cot2θ – tan2θ = cosec2θ – sec2θ
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(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
sec2θ + cosec2θ = sec2θ × cosec2θ
cot2θ - tan2θ = cosec2θ - sec2θ
`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
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 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.
सिद्ध करा:
cotθ + tanθ = cosecθ × secθ
उकल:
डावी बाजू = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
= उजवी बाजू
∴ cotθ + tanθ = cosecθ × secθ
