Representation problems

Geometry Level 3

If π 4 + tan ⁡ − 1 ( x ) \dfrac{\pi}{4} +\tan^{-1}(x) can be represented as tan ⁡ − 1 ( y ) \tan^{-1}(y)

Evaluate y \large y

Details and Assumptions

(i) x = 1 3 x = \frac{1}{3}


The answer is 2.

Kishore S. Shenoy by Kishore S. Shenoy , 22, India

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1 solution

In general arctan ⁡ ( u ) + arctan ⁡ ( v ) = arctan ⁡ ( u + v 1 − u v ) \arctan(u) + \arctan(v) = \arctan\left(\dfrac{u + v}{1 - uv}\right) for u v < 1. uv \lt 1.

In this case, since π 4 = arctan ⁡ ( 1 ) , \dfrac{\pi}{4} = \arctan(1), we have u = 1 u = 1 and v = 1 3 . v = \dfrac{1}{3}. So since u v = 1 3 < 1 uv = \dfrac{1}{3} \lt 1 we have

π 4 + arctan ⁡ ( 1 3 ) = arctan ⁡ ( 1 + 1 3 1 − 1 3 ) = arctan ⁡ ( 2 ) ⟹ y = 2 . \dfrac{\pi}{4} + \arctan\left(\dfrac{1}{3}\right) = \arctan\left(\dfrac{1 + \frac{1}{3}}{1 - \frac{1}{3}}\right) = \arctan(2) \Longrightarrow y = \boxed{2}.

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Here u v < 1 uv <1 is a weak guy...

So what if u v > 1 uv > 1 , a strong guy?

tan ⁡ − 1 ( u ) + tan ⁡ − 1 ( v ) = π + tan ⁡ − 1 [ u + v 1 − u v ] \tan^{-1}(u) + \tan^{-1}(v) = \pi + \tan^{-1} \left[\dfrac{u+v}{1-uv}\right]

Kishore S. Shenoy - 5 years, 8 months ago

I used this same solution.

Kishore S. Shenoy - 5 years, 8 months ago

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