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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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} = $

Similar Questions

The value of $\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 30^{\circ}$ is:

If $\cos x + \cos^2 x = 1$,then the value of $\sin^2 x + \sin^4 x$ is

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$ is equal to

If $f(x) = \cos^2 x + \sec^2 x$,then

If $A + B + C = \frac{\pi}{2}$,then the value of $\tan A \tan B + \tan B \tan C + \tan C \tan A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module