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

The correct option is D √(3)We have tan6π15 tanπ151+tan6π15tanπ15=tanπ3 tan6π15 tanπ15=√(3)+√(3)tan6π15tanπ15 tan6π15 tanπ15 √(3)tan6π15tanπ15=√(3) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

tan2 π/5-tanπ...

Question

$tan \frac{2 π}{5} - tan \frac{π}{15} - \sqrt{3} tan \frac{2 π}{5} tan \frac{π}{15}$ 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

The correct option is D $\sqrt{3}$
We have $\frac{tan \frac{6 π}{15} - tan \frac{π}{15}}{1 + tan \frac{6 π}{15} tan \frac{π}{15}} = tan \frac{π}{3}$
$tan \frac{6 π}{15} - tan \frac{π}{15} = \sqrt{3} + \sqrt{3} tan \frac{6 π}{15} tan \frac{π}{15}$
$tan \frac{6 π}{15} - tan \frac{π}{15} - \sqrt{3} tan \frac{6 π}{15} tan \frac{π}{15} = \sqrt{3}$

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tan2π5−tanπ15−√3tan2π5tanπ15 is equal to

The correct option is D √3We have tan6π15−tanπ151+tan6π15tanπ15=tanπ3 tan6π15−tanπ15=√3+√3tan6π15tanπ15 tan6π15−tanπ15−√3tan6π15tanπ15=√3

$tan \frac{2 π}{5} - tan \frac{π}{15} - \sqrt{3} tan \frac{2 π}{5} tan \frac{π}{15}$ is equal to