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Question
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
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Solution
डावी बाजू = sec2θ + cosec2θ
= `1/(cos^2theta) + 1/(sin^2theta)`
= `(sin^2theta + cos^2theta)/(cos^2theta*sin^2theta)`
= `1/(cos^2theta*sin^2theta)` ......[∵ sin2θ + cos2θ = 1]
= `1/(cos^2theta) xx 1/(sin^2theta)`
= sec2θ × cosec2θ
= उजवी बाजू
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
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sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= उजवी बाजू
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
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
