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# help

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Simplify tan(20) + tan(40) + sqrt(3)*tan(20)*tan(40).  (All angles are in degrees.)

Dec 13, 2019

#1
+24430
+3

Simplify

$$\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)$$.

Formula:

$$\boxed{ \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} \quad \text{or} \quad \tan(x)+\tan(y)= \Big(1-\tan(x)\tan(y)\Big)\tan(x+y) }$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) } \\\\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(20^\circ+40^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(60^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \quad | \quad \tan(60^\circ) = \sqrt{3} \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\sqrt{3}+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\sqrt{3}-\sqrt{3}\tan(20^\circ)\tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\mathbf{ \sqrt{3} } \\ \hline \end{array}$$

$$\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)=\sqrt{3}$$

Dec 13, 2019
edited by heureka  Dec 13, 2019

#1
+24430
+3

Simplify

$$\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)$$.

Formula:

$$\boxed{ \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} \quad \text{or} \quad \tan(x)+\tan(y)= \Big(1-\tan(x)\tan(y)\Big)\tan(x+y) }$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) } \\\\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(20^\circ+40^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(60^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \quad | \quad \tan(60^\circ) = \sqrt{3} \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\sqrt{3}+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\sqrt{3}-\sqrt{3}\tan(20^\circ)\tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\mathbf{ \sqrt{3} } \\ \hline \end{array}$$

$$\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)=\sqrt{3}$$

heureka Dec 13, 2019
edited by heureka  Dec 13, 2019