How to solve tan(30+60)

Welcome to my article How to solve tan(30+60). This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve tan(30+60), read and understand it carefully till the end.

Let us know how to solve this question How to solve tan(30+60)

First write the question on the page of the notebook.

How to solve tan(30+60)

Let us first write this question in this way,

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)

This question is based on trigonometry, in this way,

to solve this question, we will also take the help of trigonometry formula.



\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{\tan {{{30}}^{\circ }}+\tan {{{60}}^{\circ }}}}{{1-\tan {{{30}}^{\circ }}\times \tan {{{60}}^{\circ }}}}

we know that,

\displaystyle \tan {{30}^{\circ }}=\frac{1}{{\sqrt{3}}}

\displaystyle \tan {{60}^{\circ }}=\sqrt{3}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{\frac{1}{{\sqrt{3}}}+\sqrt{3}}}{{1-\frac{1}{{\sqrt{3}}}\times \sqrt{3}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{\frac{1}{{\sqrt{3}}}+\frac{{\sqrt{3}}}{1}}}{{1-\frac{{1\times \sqrt{3}}}{{\sqrt{3}}}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{\frac{{1+\sqrt{3}\times \sqrt{3}}}{{\sqrt{3}}}}}{{\frac{{\sqrt{3}-1\times \sqrt{3}}}{{\sqrt{3}}}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{\frac{{1+\sqrt{3}\times \sqrt{3}}}{{\cancel{{\sqrt{3}}}}}}}{{\frac{{\sqrt{3}-1\times \sqrt{3}}}{{\cancel{{\sqrt{3}}}}}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{1+\sqrt{3}\times \sqrt{3}}}{{\sqrt{3}-1\times \sqrt{3}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{{1+3}}{{\sqrt{3}-\sqrt{3}}}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\frac{4}{0}

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\infty

so that,

\displaystyle \tan \left( {{{{30}}^{\circ }}+{{{60}}^{\circ }}} \right)=\tan ({{90}^{\circ }})=\infty [Answer]

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