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Question
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
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Solution
डावी बाजू = `tanθ/(secθ + 1)`
= `tanθ/(secθ + 1) xx (secθ - 1)/(secθ - 1)` ..............[छेदाचे परिमेयकरण करून]
= `(tanθ(secθ - 1))/(sec^2θ - 1)`
= `(tanθ(secθ - 1))/(tan^2θ)` .....`[(∵ 1 + tan^2θ = sec^2θ), (∴ sec^2θ - 1 = tan^2θ)]`
= `(secθ - 1)/(tanθ)`
= उजवी बाजू
∴ `tanθ/(secθ + 1) = (secθ - 1)/tanθ`
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RELATED QUESTIONS
secθ + tanθ = `cosθ/(1 - sinθ)`
cot2θ - tan2θ = cosec2θ - sec2θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sin2θ + sin2(90 – θ) = ?
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot 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.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
जर `1/sin^2θ - 1/cos^2θ-1/tan^2θ-1/cot^2θ-1/sec^2θ-1/("cosec"^2θ) = -3`, तर θ ची किमत काढा.
