If A is an angel between 0<A<90 degree, how to prove (1+(1/sinA))(1+(1/cosA))>5.?

Question

Its a good problem that i cannt solve, so plz help me, mates.

Answer 1

multiply out to give
(1+(1/sinA))(1+(1/cosA))
= 1+1/sinA + 1/cosA + 2/sin2A

[NB sin2A = 2sinAcosA]

sinA <1 so 1/sinA >1
cosA <1 so 1/cosA > 1
sin2A < 1 so 2/sin2A > 2

therefore its >5 in total.


ps, using calculus or graphical calculator gives a min value at A=45, where the expression = 5.82842... = 3+2*sqrt2

Answer 2

Between 0 and 90 degrees,

0 < sinA < 1
0 < cosA < 1

Let's take the right hand of these inequalities.

sinA < 1
cosA < 1

If we take the reciprocal of both sides, we have to flip the inequality as well; therefore

1/sinA > 1
1/cosA > 1

If we add 1 to both sides,

1 + (1/sinA) > 2
1 + (1/cosA) > 2

And if we multiply them together,

[1 + (1/sinA)] [1 + (1/cosA)] > 4

Unfortunately I'm not sure why I came out with a 4, and the question says 5. Work in progress ...

EDIT: Aeronic, you are brilliant. Kudos to you for a nice answer.

Answer 3

Since 0
It follows then that 1 + (1/sin A) > 2 and 1 + (1/cos A) > 2. So their sum must be greater than 4.

I can't see a way for it to be definitely greater than 5, but I'll spend a little more time on it.

Answer 4

One method that you might want to try is to take the derivative of the equation and solve to find where the minimum is, then determine what that minimum value actually is.