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प्रश्न
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
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उत्तर
L.H.S. = (cosec A – sin A) (sec A – cos A) × sec2 A
= `(1/sinA - sinA)(1/cosA - cosA) xx sec^2A` ...`{∵ cosec theta = 1/sintheta, sectheta = 1/costheta, 1 - sin^2theta = cos^2theta, 1 - cos^2theta = sin^2theta}`
= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA) 1/(cos^2A)`
= `(cos^2A)/(sinA)*(sin^2A)/(cosA)*1/(cos^2A)`
= `sinA/cosA`
= tan A = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
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`cosec^4A - cosec^2A = cot^4A + cot^2A`
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`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Without using trigonometric identity , show that :
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sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
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