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Question
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
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Solution
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) = 4 sqrt ((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 * (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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