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Question
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
Solution
डावी बाजू = `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)`
= `sintheta/(1/costheta + 1) + sintheta/(1/costheta - 1`
= `sintheta/((1 + costheta)/costheta) + sintheta/((1 - costheta)/(costheta))`
= `(sintheta costheta)/(1 + costheta) + (sintheta costheta)/(1 - costheta)`
= `sin theta costheta (1 /(1 + costheta) + 1/(1 - costheta))`
= `sintheta costheta [(1 - costheta + 1 + costheta)/((1 + costheta)(1 - costheta))]`
= `sintheta costheta (2/(1 - cos^2theta))` ......[∵ (a + b)(a – b) = a2 – b2]
= `sintheta costheta xx 2/(sin^2theta)` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
= `2 xx (costheta)/(sintheta)`
= 2cot θ
= उजवी बाजू
∴ `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
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`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
खालीलपैकी चुकीचे सूत्र कोणते?
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"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
