Recall that [tex]\cos^2x+\sin^2x=1[/tex] and [tex]\sqrt{x^2}=|x|[/tex] for all [tex]x[/tex]. So
[tex]\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|[/tex]
[tex]\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|[/tex]
For [tex]0<x<\frac\pi2[/tex], we expect both [tex]\cos x>0[/tex] and [tex]\sin x>0[/tex] (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, [tex]|x|=x[/tex] if [tex]x>0[/tex].
So we have
[tex]\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}[/tex]
making H the answer.
C is always true, because the inequality reduces to x > y.
