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Question
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
Solution
डावी बाजू = (1 – cos2A) . sec2B + tan2B(1 – sin2A)
= `sin^2"A"* 1/(cos^2"B") + (sin^2"B")/(cos^2"B") (1 - sin^2"A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= `(sin^2"A")/(cos^2"B") + (sin^2"B")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B")`
= `(sin^2"A")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B") + (sin^2"B")/(cos^2"B")`
= `(sin^2"A")/(cos^2"B") (1 - sin^2"B") + tan^2"B"`
= `(sin^2"A")/(cos^2"B") (cos^2"B") + tan^2"B"`
= sin2A + tan2B
= उजवी बाजू
∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B
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sec2θ + cosec2θ = sec2θ × cosec2θ
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
cot θ + tan θ = cosec θ × sec θ, हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती:
डावी बाजू = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= उजवी बाजू
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
