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Addition Rule of Differentiation

The value of ...

Question

The value of $t a n \frac{3 π}{24} t a n \frac{5 π}{24} t a n \frac{11 π}{24}$ is equal to:

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Solution

We have,

$tan \frac{3 π}{24} tan \frac{5 π}{24} tan \frac{11 π}{24}$

$= \frac{sin \frac{3 π}{24} sin \frac{5 π}{24} sin \frac{11 π}{24}}{cos \frac{3 π}{24} cos \frac{5 π}{24} cos \frac{11 π}{24}}$

Multiplying and divide by 2 and we get.

$= \frac{2 sin \frac{3 π}{24} sin \frac{5 π}{24} sin \frac{11 π}{24}}{2 cos \frac{3 π}{24} cos \frac{5 π}{24} cos \frac{11 π}{24}}$

$= \frac{[cos (\frac{3 π}{24} - \frac{5 π}{24}) - cos (\frac{3 π}{24} + \frac{5 π}{24})] sin \frac{11 π}{24}}{[cos (\frac{3 π}{24} + \frac{5 π}{24}) + cos (\frac{3 π}{24} - \frac{5 π}{24})] cos \frac{11 π}{24}}$

$= \frac{[cos \frac{π}{12} - cos \frac{π}{3}] sin \frac{11 π}{24}}{[cos \frac{π}{3} + cos \frac{π}{12}] cos \frac{11 π}{24}}$

$= \frac{[cos \frac{π}{12} - \frac{1}{2}] sin \frac{11 π}{24}}{[\frac{1}{2} + cos \frac{π}{12}] cos \frac{11 π}{24}}$

$= \frac{[2 cos \frac{π}{12} - 1] sin \frac{11 π}{24}}{[1 + 2 cos \frac{π}{12}] cos \frac{11 π}{24}}$

$= \frac{2 cos \frac{π}{12} sin \frac{11 π}{24} - sin \frac{11 π}{24}}{cos \frac{11 π}{24} + 2 cos \frac{π}{12} cos \frac{11 π}{24}}$

$= \frac{sin (\frac{π}{12} + \frac{11 π}{24}) - sin (\frac{π}{12} - \frac{11 π}{24}) - sin \frac{11 π}{24}}{cos \frac{11 π}{24} + cos (\frac{π}{12} + \frac{11 π}{24}) + cos (\frac{π}{12} - \frac{11 π}{24})}$

$= \frac{sin (\frac{13 π}{24}) - sin (- \frac{9 π}{24}) - sin \frac{11 π}{24}}{cos \frac{11 π}{24} + cos (\frac{13 π}{24}) + cos (- \frac{9 π}{24})}$

$= \frac{sin \frac{13 π}{24} + sin \frac{9 π}{24} - sin \frac{11 π}{24}}{cos \frac{11 π}{24} + cos \frac{13 π}{24} + cos \frac{9 π}{24}}$
Hence, this is the answer.

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