Question

If $0<{\alpha }_1<{\alpha }_2<\dots \dots <{\alpha }_n<\frac{\pi }{2}$, then $tan{\alpha }_1<\frac{sin{\alpha }_1+sin{\alpha }_2+\dots \dots +sin{\alpha }_n}{cos{\alpha }_1+cos{\alpha }_2+\dots \dots +{\alpha }_n}<tan{\alpha }_n$

• A. True
• B. False

Answer 1

Answer: (A)

We know that in the interval $(0,\frac{\pi }{2})$, the function sin x is increasing and the function cos x is decreasing.
Therefore
$0<sin{\alpha }_1<sin{\alpha }_2<\dots \dots <sin{\alpha }_n$,
and $cos{\alpha }_1>cos{\alpha }_2>\dots \dots >cos{\alpha }_n>0$
Then
$nsin{\alpha }_1<sin{\alpha }_2+\dots \dots +sin{\alpha }_n\dots \dots \left(1\right)$
and $ncos{\alpha }_1>cos{\alpha }_2+\dots \dots +cos{\alpha }_n>ncos{\alpha }_n$
or $\frac{1}{ncos{\alpha }_1<\frac{1}{cos{\alpha }_1+cos{\alpha }_2+\dots \dots +cos{\alpha }_n}}<\frac{1}{ncos{\alpha }_n}\dots \dots \left(2\right)$
Now multiplying (1) and (2), we get
$tan{\alpha }_1<\frac{sin{\alpha }_1+sin{\alpha }_2+\dots \dots +sin{\alpha }_n}{cos{\alpha }_1+cos{\alpha }_2+\dots \dots +cos{\alpha }_n}<tan{\alpha }_n$