Tan2-5 -tan-15 -3tan2-5tan-15 is equal to

The correct option is D √(3) tan (π3)=tan (5π15) =tan (6π15 π15) =tan (2π5 π15) √(3)=tan (2π5) tan (π15)1+tan (2π5).tan (π15) √(3 ...

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Composition of Trigonometric Functions and Inverse Trigonometric Functions

tan2π5 -tanπ1...

Question

$t a n \frac{2 π}{5} - t a n \frac{π}{15} - \sqrt{3} t a n \frac{2 π}{5} t a n \frac{π}{15}$ is equal to

- \sqrt{3}

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\frac{1}{\sqrt{3}}

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1

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\sqrt{3}

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Solution

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1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

[\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \frac{1}{\sqrt{5} + \sqrt{4}} + . . . . + \frac{1}{\sqrt{16} + \sqrt{15}}]

\frac{(- 1 + i \sqrt{3})^{15}}{(1 - i)^{20}} + \frac{(- 1 - i \sqrt{3})^{15}}{(1 + i)^{20}}

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

A = {(4 + \sqrt{15})}^{1 / 3} + {(4 - \sqrt{15})}^{1 / 3}

A^{3} - 3 A

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