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I solved a geometry problem to the following equation $$\cot(\theta)=\sec6^\circ+ \tan36^\circ$$,it’s answer is $\theta=30^\circ$,how can I prove that $$\sec6^\circ+\tan36^\circ=\cot30^\circ=\sqrt3$$ without using a calculator.

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Xu Alain
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  • Bravo and Thank you very much. I spent almost three days trying to figure out how to solve it. I even thought about using tan 60- tan 36 but fell short to reach the final result. – Xu Alain Nov 09 '24 at 10:17

1 Answers1

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Hint:

$$\tan60^\circ-\tan36^\circ=\cdots=\dfrac{\sin(60-36)^\circ}{\cos60^\circ\cos36^\circ}$$

$$\text{ Again, }2\sin24^\circ\cos6^\circ=\sin(24+6)^\circ+\sin(24-6)^\circ =\sin30^\circ+\cos36^\circ-\dfrac12=? $$ using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

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