Find the range of $\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^{2}x}}-\frac{\cos x}{\sqrt{1+\cot^{2}x}}$
- $[0,1]$
- $[-1,0]$
- $[-2,2]$
- $\{0\}$
Simplify the expression: sin(x)/sqrt(1+tan^2(x)) = sin(x)/|sec(x)| = sin(x)cos(x)/|cos(x)|. If cos(x) > 0, this is sin(x)cos(x)/cos(x) = sin(x). If cos(x) < 0, this is -sin(x). Similarly, cos(x)/sqrt(1+cot^2(x)) = cos(x)/|csc(x)| = cos(x)sin(x)/|sin(x)|. The expression simplifies to 0 for all x where defined.
Use the trigonometric identities for secant and cosecant to simplify the denominators. The term square root of (1 + tan^2 x) is sec x, and since sec x is 1/cos x, the first term simplifies to (sin x) / (1/cos x), but considering the principal root it becomes sin x cos x for x in the first quadrant or -sin x cos x in the second. Similarly, the second term (cos x) / square root of (1 + cot^2 x) simplifies to cos x sin x. Based on the identity, the first term is sin x * |cos x| and the second term is cos x * |sin x|. However, evaluating the standard identity 1 + tan^2 x = sec^2 x implies the terms simplify to sin x cos x - cos x sin x = 0, making the range the set containing only 0.