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Problem 936 AMC 12 late, AIME early Difficulty 4.2 Find the answer Annual Harvard-MIT Mathematics Tournament · United States
Compute arctan ( tan 65 ∘ − 2 tan 40 ∘ ) \arctan \left(\tan 65^{\circ}-2 \tan 40^{\circ}\right) arctan ( tan 6 5 ∘ − 2 tan 4 0 ∘ ) 0 ∘ 0^{\circ} 0 ∘ 180 ∘ 180^{\circ} 18 0 ∘
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Official solution Solution:
Answer: 25 ∘ 25^{\circ} 2 5 ∘
First Solution: We havetan 65 ∘ − 2 tan 40 ∘ = cot 25 ∘ − 2 cot 50 ∘ = cot 25 ∘ − cot 2 25 ∘ − 1 cot 25 ∘ = 1 cot 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} . tan 6 5 ∘ − 2 tan 4 0 ∘ = cot 2 5 ∘ − 2 cot 5 0 ∘ = cot 2 5 ∘ − cot 2 5 ∘ cot 2 2 5 ∘ − 1 = cot 2 5 ∘ 1 = tan 2 5 ∘ . 25 ∘ 25^{\circ} 2 5 ∘
Second Solution: We havetan 65 ∘ − 2 tan 40 ∘ = 1 + tan 20 ∘ 1 − tan 20 ∘ − 4 tan 20 ∘ 1 − tan 2 20 ∘ = ( 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} . tan 6 5 ∘ − 2 tan 4 0 ∘ = 1 − tan 2 0 ∘ 1 + tan 2 0 ∘ − 1 − tan 2 2 0 ∘ 4 tan 2 0 ∘ = ( 1 − tan 2 0 ∘ ) ( 1 + tan 2 0 ∘ ) ( 1 − tan 2 0 ∘ ) 2 = tan ( 4 5 ∘ − 2 0 ∘ ) = tan 2 5 ∘ . 25 ∘ 25^{\circ} 2 5 ∘
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