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If \( \Large \tan A + \sin A = m\ and\ \tan A - \sin A = n,\ then\ \frac{ \left(m^{2}-n^{2}\right)^{2} }{mn} \) is equal to:
|
A) 4 |
|
B) 3 |
|
C) 16 |
|
D) 9 |
Correct Answer:
| C) 16 |
Description for Correct answer:
Since, \( \Large \tan A + \sin A = m \)
and \( \Large \tan A - \sin A = n \)
\( \Large \therefore m+n = 2 \tan A \)
and \( \Large m-n = 2 \sin A \)
Also \( \Large mn = \left(\tan A + \sin A\right) \left(\tan A - \sin A\right) \)
= \( \Large \tan^{2}A - \sin^{2}A \)
Now \( \Large \frac{m^{2}-n^{2}}{mn} = \frac{ \left(m+n\right)^{2} \left(m-n\right)^{2} }{mn} \)
=\( \Large \frac{ \left(2 \tan A\right)^{2} \left(2 \sin A\right)^{2} }{\tan^{2}A-\sin^{2}A} \)
= \( \Large \frac{16 \tan^{2}A \sin^{2}A}{\sin^{2}A \tan^{2}A} = 16 \)
Since, \( \Large \tan A + \sin A = m \)
and \( \Large \tan A - \sin A = n \)
\( \Large \therefore m+n = 2 \tan A \)
and \( \Large m-n = 2 \sin A \)
Also \( \Large mn = \left(\tan A + \sin A\right) \left(\tan A - \sin A\right) \)
= \( \Large \tan^{2}A - \sin^{2}A \)
Now \( \Large \frac{m^{2}-n^{2}}{mn} = \frac{ \left(m+n\right)^{2} \left(m-n\right)^{2} }{mn} \)
=\( \Large \frac{ \left(2 \tan A\right)^{2} \left(2 \sin A\right)^{2} }{\tan^{2}A-\sin^{2}A} \)
= \( \Large \frac{16 \tan^{2}A \sin^{2}A}{\sin^{2}A \tan^{2}A} = 16 \)
Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio
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