Answer to Question #3109 in Trigonometry for Sandeepan dhar chowdhury

Question

Q. Prove the identity:
& 1 - (sin[sup]2[/sup]x/(1+cotx)) - (cos[sup]2[/sup]x/(1+tanx)) = sin x cos x

Answer 1

<img src="/cgi-bin/mimetex.cgi?1-%20%5Cfrac%7B%5Csin%5E2%20x%7D%7B1%20+%20%5Ccot%20x%7D%20-%20%5Cfrac%7Bcos%5E2%20x%7D%7B1%20+%20%5Ctan%20x%7D%20=%201%20-%20%5Cfrac%7B%5Csin%5E3%20x%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20-%20%5Cfrac%7Bcos%5E3%20x%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20=%20%5C%5C%20=%20%5Cfrac%7B%5Csin%20x%20+%20%5Ccos%20x%20-%20%5Csin%5E3%20x%20-%20%5Ccos%5E3%20x%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20=%20%5Cfrac%7B%5Csin%20x%20%28%201-%20%5Csin%5E2%20x%29%20+%20%5Ccos%20x%20%281%20-%20%5Ccos%5E2%20x%29%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20=%20%5C%5C%20=%20%5Cfrac%7B%5Csin%20x%20%5Ccos%5E2%20x%20+%20%5Ccos%20x%20%5Csin%5E2%20x%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20=%20%5Cfrac%7B%5Csin%20x%20%5Ccos%20x%20%28%5Csin%20x%20+%20%5Ccos%20x%29%7D%7B%5Csin%20x%20+%20%5Ccos%20x%7D%20=%20%5Csin%20x%20%5Ccos%20x" title="1- \frac{\sin^2 x}{1 + \cot x} - \frac{cos^2 x}{1 + \tan x} = 1 - \frac{\sin^3 x}{\sin x + \cos x} - \frac{cos^3 x}{\sin x + \cos x} = \\ = \frac{\sin x + \cos x - \sin^3 x - \cos^3 x}{\sin x + \cos x} = \frac{\sin x ( 1- \sin^2 x) + \cos x (1 - \cos^2 x)}{\sin x + \cos x} = \\ = \frac{\sin x \cos^2 x + \cos x \sin^2 x}{\sin x + \cos x} = \frac{\sin x \cos x (\sin x + \cos x)}{\sin x + \cos x} = \sin x \cos x">