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प्रश्न
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
उत्तर
डावी बाजू = `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")`
= `((1 +sin "B")^2 + cos^2"B")/(cos "B"(1 + sin "B"))`
= `(1 +2sin"B" + sin^2"B" + cos^2"B")/(cos"B"(1 + sin"B"))` ......[∵ (a + b)2 = a2 + 2ab + b2]
= `(1 + 2sin"B" + 1)/(cos"B"(1+ sin"B"))` .....[∵ sin2B + cos2B = 1]
= `(2 + 2sin"B")/(cos"B"(1 + sin"B"))`
= `(2(1 + sin"B"))/(cos"B"(1 + sin"B"))`
= `2/"cos B"`
= 2 sec B
= उजवी बाजू
∴ `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
APPEARS IN
संबंधित प्रश्न
`(sin^2θ)/(cosθ) + cosθ = secθ`
cos2θ(1 + tan2θ) = 1
secθ + tanθ = `cosθ/(1 - sinθ)`
tan4θ + tan2θ = sec4θ - sec2θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
