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Question
जर 1 – cos2θ = `1/4`, तर θ = ?
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Solution
1 – cos2θ = `1/4` ......[दिलेले]
∴ sin2θ = `1/4` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
∴ sin θ = `1/2` ......[दोन्ही बाजूंचे वर्गमूळ घेऊन]
∴ θ = 30° ......`[because sin 30^circ = 1/2]`
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RELATED QUESTIONS
cos2θ(1 + tan2θ) = 1
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
sec6x - tan6x = 1 + 3sec2x × tan2x
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर cos θ = `24/25`, तर sin θ = ?
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.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
θ चे निरसन करा:
जर x = r cosθ आणि y = r sinθ
