First slide

Trigonometric Identities

Question

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

Moderate

A

4
B

3
C

16
D

9

Solution

Since,
$\begin{matrix} \tan A + \sin A = m \\ and \tan A - \sin A = n \\ ∴ m + n = 2 \tan A \\ and m - n = 2 \sin A \\ Also, mn = (\tan A + \sin A) (\tan A - \sin A) \\ = \tan^{2} A - \sin^{2} A \\ Now, \frac{{(m^{2} - n^{2})}^{2}}{mn} = \frac{(m + n)^{2} (m - n)^{2}}{mn} \\ = \frac{(2 \tan A)^{2} (2 \sin A)^{2}}{\tan^{2} A - \sin^{2} A} \\ = \frac{16 \tan^{2} {Asin}^{2} A}{\sin^{2} {Atan}^{2} A} = 16 \end{matrix}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App