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Question
1 + tan2θ = किती?
Options
cot2θ
cosec2θ
sec2θ
tan2θ
MCQ
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Solution
1 + tan2θ = sec2θ
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त्रिकोणमितीय नित्यसमानता
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cot θ + tan θ = cosec θ sec θ
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
sec θ(1 - sin θ) (sec θ + tan θ) = 1
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
