Maths Olympiad Prep

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Geometry Difficulty 4.2 AIME Find the answer United States

Problem:

Compute arctan⁡(tan⁡65∘−2tan⁡40∘)\arctan \left(\tan 65^{\circ}-2 \tan 40^{\circ}\right). (Express your answer in degrees as an angle between 0∘0^{\circ} and 180∘180^{\circ}.)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Answer: 25∘25^{\circ}

First Solution: We have
tan⁡65∘−2tan⁡40∘=cot⁡25∘−2cot⁡50∘=cot⁡25∘−cot⁡225∘−1cot⁡25∘=1cot⁡25∘=tan⁡25∘. \tan 65^{\circ}-2 \tan 40^{\circ}=\cot 25^{\circ}-2 \cot 50^{\circ}=\cot 25^{\circ}-\frac{\cot ^{2} 25^{\circ}-1}{\cot 25^{\circ}}=\frac{1}{\cot 25^{\circ}}=\tan 25^{\circ} .
Therefore, the answer is 25∘25^{\circ}.

Second Solution: We have
tan⁡65∘−2tan⁡40∘=1+tan⁡20∘1−tan⁡20∘−4tan⁡20∘1−tan⁡220∘=(1−tan⁡20∘)2(1−tan⁡20∘)(1+tan⁡20∘)=tan⁡(45∘−20∘)=tan⁡25∘. \tan 65^{\circ}-2 \tan 40^{\circ}=\frac{1+\tan 20^{\circ}}{1-\tan 20^{\circ}}-\frac{4 \tan 20^{\circ}}{1-\tan ^{2} 20^{\circ}}=\frac{\left(1-\tan 20^{\circ}\right)^{2}}{\left(1-\tan 20^{\circ}\right)\left(1+\tan 20^{\circ}\right)}=\tan \left(45^{\circ}-20^{\circ}\right)=\tan 25^{\circ} .
Again, the answer is 25∘25^{\circ}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.