If tan A + sin A = m and tan A - sin A = n, t | vedclass

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(C) Given: $m = 	an A + \sin A$ and $n = 	an A - \sin A$.First,calculate $m^2 - n^2$:$m^2 - n^2 = (m + n)(m - n) = (2 	an A)(2 \sin A) = 4 	an A \

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given: $m = 	an A + \sin A$ and $n = 	an A - \sin A$.First,calculate $m^2 - n^2$:$m^2 - n^2 = (m + n)(m - n) = (2 	an A)(2 \sin A) = 4 	an A \sin A$.Next,calculate $mn$:$mn = (	an A + \sin A)(	an A - \sin A) = 	an^2 A - \sin^2 A = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \sin^2 A \left( \frac{1 - \cos^2 A}{\cos^2 A} ight) = \sin^2 A \cdot \frac{\sin^2 A}{\cos^2 A} = 	an^2 A \sin^2 A$.Now,substitute these into the expression $\frac{(m^2 - n^2)^2}{mn}$:$\frac{(4 	an A \sin A)^2}{	an^2 A \sin^2 A} = \frac{16 	an^2 A \sin^2 A}{	an^2 A \sin^2 A} = 16$.

If $\tan A + \sin A = m$ and $\tan A - \sin A = n,$ then $\frac{(m^2 - n^2)^2}{mn} = $

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