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If ${tan}^{2} 30^{\circ} + {tan}^{2} 45^{\circ} + {tan}^{2} 60^{\circ}$ is $m \frac{1}{3}$ , then $m$ is:

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Solution

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${tan}^{2} 30^{\circ} + {tan}^{2} 45^{\circ} + {tan}^{2} 60^{\circ}$
= $(\frac{1}{\sqrt{3}})^{2} + 1^{2} + (\sqrt{3})^{2}$

= $\frac{1}{3} + 1 + 3$

= $\frac{1 + 3 + 9}{3}$

= $\frac{13}{3}$

= $4 \frac{1}{3}$

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