যদি ∠P+∠Q = 90° হয়, তবে দেখাই যে, \(\sqrt{\cfrac{\sin P}{\cos Q}-\sin P \cos Q}=\cos P\)
\(\sqrt{\cfrac{\sin P}{\cos Q}-\sin P \cos Q}\)
\(=\sqrt{\cfrac{\sin P}{\cos (90°-P)}-\sin P \cos (90°-P)}\)
\(=\sqrt{\cfrac{\sin P}{\sin P}-\sin P \sin P}\)
\(=\sqrt{\cfrac{\sin P}{\sin P}-\sin^2 P }\)
\(=\sqrt{1-\sin^2 P }\)
\(=\sqrt{\cos^2 P }\)
\(=cosP\) (প্রমানিত)