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प्रश्न
cot θ + tan θ = cosec θ sec θ
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उत्तर
डावी बाजू = cot θ + tan θ
= `cos θ/sin θ + sin θ/cos θ`
= `(cos^2 θ + sin^2 θ)/(sin θcos θ)`
= `1/(sin θcos θ)` .....[∵ sin2θ + cos2θ = 1]
= `1/sinθ . 1/cosθ`
= cosec θ . sec θ
= उजवी बाजू
∴ cot θ + tan θ = cosec θ sec θ
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संबंधित प्रश्न
`(sin^2θ)/(cosθ) + cosθ = secθ`
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
cot2θ - tan2θ = cosec2θ - sec2θ
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
(sec θ + tan θ) . (sec θ – tan θ) = ?
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
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
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
