Trigonometric Proof: Question:
If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$
My Efforts:
$(m\cos\alpha-n\sin\alpha)^{2}=p^2$
$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$
Now i think we need to add something on both side but i can't figure out what to add
 A: HINT:
$$(m\cos\alpha-n\sin\alpha)^2+(m\sin\alpha+n\cos\alpha)^2=?$$
See also : Brahmagupta–Fibonacci identity
A: $$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
multiply by $-1$ and add $m^2+n^2$ to both sides :
$$m^2+n^2-m^2\cos^2\alpha-n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(1-\cos^2\alpha)+n^2(1-\sin^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(\sin^2\alpha)+n^2(\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$(m\sin\alpha+n\cos\alpha)^2=m^2+n^2-p^2$$
$$m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$$
A: From 
$$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
we can add $m^2\sin^2\alpha+n^2\cos^2\alpha$ to both sides (to make use of the $\sin^2x+\cos^2x=1$ identity) to achieve 
$$m^2(\sin^2\alpha+\cos^2\alpha)+n^2(\sin^2\alpha+\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=p^2+m^2\sin^2\alpha+n^2\cos^2\alpha
\\\iff m^2+n^2-p^2=m^2\sin^2\alpha+n^2\cos^2\alpha+2mn\cos\alpha\ \sin\alpha=(m\sin\alpha+n\cos\alpha)^2$$
The result follows.
