This is not hard

Geometry Level 2

The right triangle above has side lengths a a and b b such that 0 < b < a 0 < b < a . One of its angles measures tan ⁡ − 1 ( a b ) \tan^{-1} \left(\frac ab\right) . What is cos ⁡ ( tan ⁡ − 1 ( a b ) ) \cos \left(\tan^{-1} \left(\frac ab \right) \right) ?

a b \frac ab b a \frac ba a 2 + b 2 a \frac {\sqrt{a^2+b^2}}a a a 2 + b 2 \frac a{\sqrt{a^2+b^2}} b a 2 + b 2 \frac b{\sqrt{a^2+b^2}}

Ricky Huang by Ricky Huang , 17, China

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1 solution

Let the angle be θ \theta .

tan ⁡ θ = a b \tan \theta = \dfrac{a}{b}

⟹ θ is the upper angle. \implies \theta \text{ is the upper angle.}

cos ⁡ θ = adjacent hypotenuse = b a 2 + b 2 \cos \theta =\dfrac{\text{adjacent}}{\text{hypotenuse}}=\boxed{\dfrac{b}{\sqrt{a^2+b^2}}}

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