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प्रश्न
If tan A + sin A = m and tan A - sin A = n, then show that m2 - n2 = 4 `sqrt(mn)`.
उत्तर
Here,
m2 - n2 = (tan A + sin A)2 - (tan A - sin A)2
m2 - n2 = ( tan A + sin A + tan A - sin A )( tan A + sin A - tan A + sin A)
m2 - n2 = (2 tan A)(2 sin A)
m2 - n2 = 4 tan A sin A ....(1)
Also,
`4 sqrt(mn) = 4sqrt((tan A + sin A)( tan A - sin A))`
= 4 `sqrt(tan^2 A - sin^2 A)`
= 4 `sqrt((sin^2 A)/(cos^2 A) - sin^2 A)`
= `4 sin A sqrt((1 - cos^2 A)/(cos^2 A))`
= `4 sin A sqrt((sin^2 A)/(cos^2 A))`
= `4 sin A sqrt((sin A)/(cos A))`
= 4 sin A. tan A ....(2)
Using equation (1) and equation (2) we get the required conditions.
Hence proved.
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