Question

If $\tan \theta +\sin \theta =m,\tan \theta -\sin \theta =n$, then $(m^2-n^2{)}^2=$.

• A. $16mn$
• B. $4mn$
• C. $32mn$
• D. $8mn$

Answer 1

The correct option is A $16mn$

we solve to get
$\tan \theta =\frac{m+n}{2}$ and $\sin \theta =\frac{m-n}{2}$
$\therefore \cot \theta =\frac{2}{m+n}$ and $cosec\theta =\frac{2}{m-n}$
now we use ${cosec}^2\theta -{\cot }^2\theta =1$ to eliminate $\theta$

$\frac{4}{(m-n{)}^2}-\frac{4}{(m+n{)}^2}=1$

$4\left[\right(m+n{)}^2-(m-n{)}^2]=(m^2-n^2{)}^2$

$4\left(4mn\right)=(m^2-n^2{)}^2$