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प्रश्न
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
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उत्तर
डावी बाजू = sec2θ + cosec2θ
= `1/(cos^2theta) + 1/(sin^2theta)`
= `(sin^2theta + cos^2theta)/(cos^2theta*sin^2theta)`
= `1/(cos^2theta*sin^2theta)` ......[∵ sin2θ + cos2θ = 1]
= `1/(cos^2theta) xx 1/(sin^2theta)`
= sec2θ × cosec2θ
= उजवी बाजू
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
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संबंधित प्रश्न
`(sin^2θ)/(cosθ) + cosθ = secθ`
secθ + tanθ = `cosθ/(1 - sinθ)`
sec2θ + cosec2θ = sec2θ × cosec2θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ 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 cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
