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Algebra Difficulty 5.0 AIME Find the answer

 17. tan⁡24∘+3tan⁡24∘⋅tan⁡36∘+tan⁡36∘ is  \text { 17. } \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \text { is }

the value of \qquad .

A number or a short expression. Spacing and $ signs are ignored.

Solution

 17. 3. tan⁡24∘+3tan⁡24∘⋅tan⁡36∘+tan⁡36∘=tan⁡(24∘+36∘)(1−tan⁡24∘⋅tan⁡36∘)+3tan⁡24∘⋅tan⁡36∘=3−3tan⁡24∘⋅tan⁡36∘+3tan⁡24∘⋅tan⁡36∘=3.\begin{array}{l}\text { 17. } \sqrt{3} \text {. } \\ \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \\ =\tan \left(24^{\circ}+36^{\circ}\right)\left(1-\tan 24^{\circ} \cdot \tan 36^{\circ}\right)+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ} \\ =\sqrt{3}-\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}=\sqrt{3} .\end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.