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`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
Concept: undefined >> undefined
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
Concept: undefined >> undefined
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`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
Concept: undefined >> undefined
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
Concept: undefined >> undefined
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
Concept: undefined >> undefined
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
Concept: undefined >> undefined
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
Concept: undefined >> undefined
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
Concept: undefined >> undefined
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
Concept: undefined >> undefined
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
Concept: undefined >> undefined
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
Concept: undefined >> undefined
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
Concept: undefined >> undefined
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
Concept: undefined >> undefined
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
Concept: undefined >> undefined
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
Concept: undefined >> undefined
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
Concept: undefined >> undefined
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.
Concept: undefined >> undefined
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
Concept: undefined >> undefined
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
Concept: undefined >> undefined
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
Concept: undefined >> undefined
