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प्रश्न
(sec A + tan A) (1 − sin A) = ______.
पर्याय
sec A
sin A
cosec A
cos A
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उत्तर
(sec A + tan A) (1 − sin A) = cos A.
Explanation:
The given expression is `(sec "A"+tan "A") (1-sin "A")`.
Simplifying the given expression, we have
`(sec "A"+tan "A")(1-sin "A")`
= `(1/cos "A"+sin "A"/cos "A")(1-sin "A")`
= `(1+sin "A")/(cos"A")xx(1-sin "A")`
= `((1+sin "A")(1-sin "A"))/(cos "A")`
= `(1-sin^2 "A")/cos "A"`
= `cos^2 "A"/cos "A"`
= `cos "A"`
संबंधित प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
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`(tan A + tan B)/(cot A + cot B) = tan A tan B`
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
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`cosecA + cotA = 1/(cosecA - cotA)`
Prove that:
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`cot^2A/(cosecA - 1) - 1 = cosecA`
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sec4 A − sec2 A is equal to
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secA(1 - sinA)(secA + tanA) = 1
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Choose the correct alternative:
cos θ. sec θ = ?
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
