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Question
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
Solution
डावी बाजू = `sec"A"/(tan "A" + cot "A")`
= `sec"A"/((sin"A")/(cos"A") + (cos"A")/(sin"A"))`
= `sec"A"/((sin^2"A" + cos^2"A")/(cos"A" sin"A"))`
= `sec"A"/(1/(cos"A" sin"A"))` ......[∵ sin2A + cos2A = 1]
= sec A cos A sin A
= `1/cos"A" xx cos "A" sin "A"`
= sin A
= उजवी बाजू
∴ `sec"A"/(tan "A" + cot "A")` = sin A
APPEARS IN
RELATED QUESTIONS
cot θ + tan θ = cosec θ sec θ
sec4θ - cos4θ = 1 - 2cos2θ
sec4A(1 - sin4A) - 2tan2A = 1
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sec2θ – tan2θ = ?
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
जर tan θ – sin2θ = cos2θ, तर sin2θ = `1/2` हे दाखवा.
