Question

Can someone solve this?

Answer 1

• First question:

Recall that
`\cos^2x+\sin^2x=1` and
`√(x^2)=|x|` for all
`x`. So

`√(1-\cos^2x)=√(\sin^2x)=|\sin x|`

`√(1-\sin^2x)=√(\cos^2x)=|\cos x|`

For
`0<x<\frac\pi2`, 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

`(√(1-\cos^2x))/(\sin x)+(√(1-\sin^2x))/(\cos x)=(\sin x)/(\sin x)+(\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.