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Algebra of Derivatives

tanπ/24tan3π/...

Question

$t a n \frac{π}{24} t a n \frac{3 π}{24} t a n \frac{5 π}{24} t a n \frac{7 π}{24} t a n \frac{9 π}{24} t a n \frac{11 π}{24} t a n \frac{16 π}{24}$

A

1

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B

-1

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C

\sqrt{3}

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D

- \sqrt{3}

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Solution

The correct option is D $- \sqrt{3}$
$tan \frac{π}{24} tan \frac{3 π}{24} tan \frac{5 π}{24} tan \frac{7 π}{24} tan \frac{9 π}{24} tan \frac{11 π}{24} tan \frac{16 π}{24}$ ---- ( 1 )
We know,
$tan θ = cot (90 - θ)$
Now,
$tan \frac{π}{24} = cot (\frac{π}{2} - \frac{π}{24})$

$= cot (\frac{(12 - 1) π}{24})$

$= cot \frac{11 π}{24}$

$tan \frac{3 π}{24} = cot (\frac{π}{2} - \frac{3 π}{24})$

$= cot (\frac{12 π - 3 π}{24})$

$= cot \frac{9 π}{24}$

$tan \frac{5 π}{24} = cot (\frac{π}{2} - \frac{5 π}{24})$

$= cot (\frac{12 π - 5 π}{24})$

$= cot \frac{7 π}{24}$

Substituting above values in ( 1 ),
$tan \frac{π}{24} tan \frac{3 π}{24} tan \frac{5 π}{24} tan \frac{7 π}{24} tan \frac{9 π}{24} tan \frac{11 π}{24} tan \frac{16 π}{24}$

$\Rightarrow$ $cot \frac{11 π}{24} cot \frac{9 π}{24} cot \frac{7 π}{24} tan \frac{7 π}{24} tan \frac{9 π}{24} tan \frac{11 π}{24} tan \frac{16 π}{24}$

$\Rightarrow$ $(cot \frac{11 π}{24} tan \frac{11 π}{24}) (cot \frac{9 π}{24} tan \frac{9 π}{24}) (cot \frac{7 π}{24} tan \frac{7 π}{24}) tan \frac{16 π}{24}$

We know, $tan θ . cot θ = 1$

$\Rightarrow$ $(1) . (1) . (1) . tan \frac{16 π}{24}$

$\Rightarrow$ $tan \frac{2 π}{3}$

$\Rightarrow$ $\frac{sin \frac{2 π}{3}}{cos \frac{2 π}{3}}$

$\Rightarrow$ $\frac{\frac{\sqrt{3}}{2}}{\frac{- 1}{2}}$

$\Rightarrow$ $- \sqrt{3}$

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Similar questions

Q. The value of

t a n \frac{3 π}{24} t a n \frac{5 π}{24} t a n \frac{11 π}{24}

tan (\frac{π}{3} + x)

tan (\frac{π}{3} - x) = \frac{2 cos 2 x + 1}{2 cos 2 x - 1}

Q. The value of the expression

\frac{(1 + tan \frac{π}{6}) (1 - cot \frac{π}{6})}{(1 + cos \frac{π}{3}) (1 - sec \frac{π}{3})}

tan $\frac{π}{3}$ = $\sqrt{3}$

x = {tan}^{- 1} (tan \frac{5 π}{4})

y = {tan}^{- 1} (tan \frac{π}{3})

, then show that:

4 x = 3 y

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