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प्रश्न
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
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उत्तर
L.H.S. = `sinAcosA - (sinAsinAcosA)/(cosecA) - (cosAcosAsinA)/secA`
= sin A cos A – sin2 A cos A sin A – cos2 A sin A cos A
= sin A cos A – sin3 A cos A – cos3 A sin A
= sin A cos A [1 – sin2 A – cos2 A]
= sin A cos A [1 – (sin2 A + cos2 A)]
= sin A cos A (1 – 1)
= sin A cos A × 0
= 0 = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove the following identity :
sinθcotθ + sinθcosecθ = 1 + cosθ
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Prove the following identities.
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`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
