Tan6-9-33 tan4-9-27 tan2-9-

Tan 3θ=3 tan θ tan^3θ/1 3 tan^23θ⇒√(3)=tan 3×π/9=3 tanπ/9 tan^3π/9/1 3 tan^2π/9 ⇒ [ √(3) ( 1 3tan^2π/9 ) ]^2= [ 3 tanπ/9 tan^3π/9 ]^2⇒ tan^6π/9 33 tan^4π/9+27 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

tan6π/9-33 ta...

Question

$t a n^{6} \frac{π}{9} - 33 t a n^{4} \frac{π}{9} + 27 t a n^{2} \frac{π}{9}$ =

t a n \frac{π}{3}

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t a n^{2} \frac{π}{3}

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t a n \frac{π}{6}

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t a n^{2} \frac{π}{6}

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Solution

The correct option is B $t a n^{2} \frac{π}{3}$
$t a n 3 θ = \frac{3 t a n θ - t a n^{3} θ}{1 - 3 t a n^{2} 3 θ} \Rightarrow \sqrt{3} = t a n 3 \times \frac{π}{9} = \frac{3 t a n \frac{π}{9} - t a n^{3} \frac{π}{9}}{1 - 3 t a n^{2} \frac{π}{9}}$
$\Rightarrow {[\sqrt{3} (1 - 3 t a n^{2} \frac{π}{9})]}^{2} = {[3 t a n \frac{π}{9} - t a n^{3} \frac{π}{9}]}^{2} \Rightarrow t a n^{6} \frac{π}{9} - 33 t a n^{4} \frac{π}{9} + 27 t a n^{2} \frac{π}{9} = 3$
$= t a n^{2} \frac{π}{3}$

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tan6π9−33 tan4π9+27 tan2π9=

The correct option is B tan2π3tan 3θ=3 tan θ−tan3θ1−3 tan23θ⇒√3=tan 3×π9=3 tanπ9−tan3π91−3 tan2π9 ⇒[√3(1−3tan2π9)]2=[3 tanπ9−tan3π9]2⇒tan6π9−33 tan4π9+27tan2π9=3 ​=tan2π3

$t a n^{6} \frac{π}{9} - 33 t a n^{4} \frac{π}{9} + 27 t a n^{2} \frac{π}{9}$ =