Question

Can someone solve this?

Answer 1

• First question:

Recall that $\cos^2x+\sin^2x=1$ and $\sqrt{x^2}=|x|$ for all $x$. So

$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|$

$\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|$

For $0$, we expect both $\cos x>0$ and $\sin x>0$ (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, $|x|=x$ if $x>0$.

So we have

$\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}$

making H the answer.

• Second question:

C is always true, because the inequality reduces to x > y.