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प्रश्न
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
उत्तर
डावी बाजू = `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1)`
= `(cot"A" + "cosec A" - ("cosec"^2"A" - cot^2"A"))/(cot"A" - "cosec A" + 1)` .....`[(because 1 + cot^2"A" = "cosec"^2"A"),(therefore "cosec"^2"A" - cot^2"A" = 1)]`
= `(cot"A" + "cosec A" - ("cosec A" + cot"A")("cosec A" - cot"A"))/(cot"A" - "cosec A" + 1)` .....[∵ a2 – b2 = (a + b) (a – b)]
= `((cot"A" + "cosec A")(1 - "cosec A" + cot "A"))/(cot"A" - "cosec A" + 1)`
= cot A + cosec A
= `"cos A"/"sin A" + 1/"sin A"`
= `(cos "A" + 1)/"sin A"`
= उजवी बाजू
∴ `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
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संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
sec4A(1 - sin4A) - 2tan2A = 1
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(sec θ + tan θ) . (sec θ – tan θ) = ?
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
sin6A + cos6A = 1 – 3sin2A . cos2A 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
sin2θ + cos2θ ची किंमत काढा.
उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
