Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n,$ then which of the following relation between $m,n$ is correct?

• A. $m^2-n^2=4mn$
• B. $m^2+n^2=4mn$
• C. $m^2-n^2=m^2+n^2$
• D. $m^2-n^2=4\sqrt{mn}$

Answer 1

The correct option is D $m^2-n^2=4\sqrt{mn}$

Given:
$\tan \theta +\sin \theta =m\tan \theta -\sin \theta =n$
$m+n=2\tan \theta \Rightarrow \cot \theta =\frac{2}{m+n}m-n=2\sin \theta \Rightarrow \text{cosec}\theta =\frac{2}{m-n}$
Now, we know that
${\text{cosec}}^2\theta -{\cot }^2\theta =1\Rightarrow \frac{4}{(m-n{)}^2}-\frac{4}{(m+n{)}^2}=1\Rightarrow \frac{(m+n{)}^2-(m-n{)}^2}{(m^2-n^2{)}^2}=\frac{1}{4}\Rightarrow \frac{4mn}{(m^2-n^2{)}^2}=\frac{1}{4}\Rightarrow (m^2-n^2{)}^2=16mn\Rightarrow m^2-n^2=\pm 4\sqrt{mn}$

Hence, from the given options,
$m^2-n^2=4\sqrt{mn}$