Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$,then
prove that $m^2-n^2=4\sin \theta \tan \theta$.

Answer 1

Given:

$\tan \theta +\sin \theta =m\dots \left(1\right)$
$\tan \theta -\sin \theta =n\dots \left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get
$m+n=2\tan \theta \dots \left(3\right)$

Subtracting equation $\left(2\right)$ from $\left(1\right)$, we get
$m-n=2\sin \theta \dots \left(4\right)$

Multiplying $\left(3\right)$ and $\left(4\right)$, we get
$4\sin \theta \tan \theta =(m+n)(m-n)$
$\Rightarrow m^2-n^2=4\sin \theta \tan \theta$
$[\because (a+b\left)\right(a-b)=a^2-b^2]$

Hence, Proved.