Rational Inequality?

Algebra Level 3

$\frac{1}{a^2 + 1 } + \frac{1}{ b^2 + 1 } = 1$

If $a$ and $b$ are positive reals that satisfy the above equation, what can we say about the value $ab$ ?

It can be greater than 1, and can be less than 1 It is always equal to 1 It can be greater than 1, but cannot be less than 1 It can be less than 1, but cannot be greater than 1

1 solution

Kushal Bose
Oct 27, 2016

Let $\frac{1}{a^2+1} =\sin^2{x}$ and $\frac{1}{b^2+1}=\cos^2{x}$

Then $a=|cotx|$ and $b=|tanx|$ .So, $ab= 1$

Wow, that's such an interesting way of looking at it!

I had to multiply everything out to get $(ab)^2 = 1$ , thanks for showing this cute approach.

Chung Kevin - 4 years, 7 months ago

Why was the answer changed?

Shaun Leong - 4 years, 7 months ago

I initially forgot the word positive, so I added it in and then asked them to mark the "it can be less than 1, but cannot be greater than 1" as correct.

Calvin forgot to update the answer back to "It is equal to 1", and he just did it a few hours ago.

Chung Kevin - 4 years, 7 months ago

Rational Inequality?

1 solution

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