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प्रश्न
(sec A + tan A) (1 − sin A) = ______.
विकल्प
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 the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
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`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
If `sec theta + tan theta = x," find the value of " sec theta`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
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`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
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`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
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Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
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If 1 – cos2θ = `1/4`, then θ = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
