Trigonometry

we have $m = \cot\theta +\cos\theta$ and $n= \cot\theta - \cos\theta$ , we need to find the value of $(m^{2}-n^{2})^2$ ?

Notice $m+n= 2\cot\theta$ and $m-n= 2\cos\theta$

Thus: $\large\frac{m+n}{2}$ = $cot\theta$ and

$\large\frac{m-n}{2}$ = $\cos\theta$ .

Now $\sin\theta$ = $\large\frac{\cos\theta}{\cot\theta}$ = $\large\frac{m-n}{m+n}$ .

By Pythagoras, we know that $\large sin^{2}(x)+cos^{2}(x)=1$ $\implies$ ( $\large\frac{m-n}{2}$ ) $^{2}$ +( $\large\frac{m-n}{m+n}$ ) $^{2}$ = $1$

Multiply all the terms by $4(m+n)^{2}$ $\implies$ $4(m+n)^{2}$ = $4(m-n)^{2}$ + $(m+n)^{2}$ ( $m-n)^{2}$

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

$\implies$ ( $\large\ m^{2}-n^{2}$ ) $\large^{2}$ = $\large4[ (m+n)^{2} - (m-n)^{2} ]$ = $\large4[ m^{2} +2mn +n^{2} - m^{2} + 2mn -n^{2} ]$

$\implies$ ( $\large\ m^{2}-n^{2}$ ) $\large^{2}$ = $\large\ 4 * 4mn$ = $\large 16mn$