Question

Hey guyz,

Solve this question if u can....

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Answer 1

Step-by-step explanation:

We have,

$\tan( \theta ) = 2 - \sqrt{3}$

So,

$\cot(\theta) = \frac{1}{ \tan(\theta) } = \frac{1}{2 - \sqrt{3} }$

$= > \cot( \theta) = \frac{2 + \sqrt{3} }{(2 - \sqrt{3})(2 + \sqrt{3} ) }$

$= > \cot( \theta) = 2 + \sqrt{3}$

Now,

$\tan^{3} (\theta) + \cot^{3} (\theta) - 2 = (2 - \sqrt{3} )^{3} + (2 + \sqrt{3} )^{3} - 2$

We know , (a + b)³ + (a - b)³ = 2( a³ + 3ab² )

So, we have,

$\tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 2((2)^{3} + 3(2)( \sqrt{3} )^{2} )$

$= > \tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 2(8 + 18)$

$= > \tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 52$

Answer 2

Step-by-step explanation:

We have,

\tan( \theta ) = 2 - \sqrt{3}tan(θ)=2−3

So,

\cot(\theta) = \frac{1}{ \tan(\theta) } = \frac{1}{2 - \sqrt{3} }cot(θ)=tan(θ)1=2−31

= > \cot( \theta) = \frac{2 + \sqrt{3} }{(2 - \sqrt{3})(2 + \sqrt{3} ) }=>cot(θ)=(2−3)(2+3)2+3

= > \cot( \theta) = 2 + \sqrt{3}=>cot(θ)=2+3

Now,

\tan^{3} (\theta) + \cot^{3} (\theta) - 2 = (2 - \sqrt{3} )^{3} + (2 + \sqrt{3} )^{3} - 2tan3(θ)+cot3(θ)−2=(2−3)3+(2+3)3−2

We know , (a + b)³ + (a - b)³ = 2( a³ + 3ab² )

So, we have,

\tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 2((2)^{3} + 3(2)( \sqrt{3} )^{2} )tan3(θ)+cot3(θ)−2=2((2)3+3(2)(3)2)

= > \tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 2(8 + 18)=>tan3(θ)+cot3(θ)−2=2(8+18)

= > \tan^{3} (\theta) + \cot^{3} (\theta) - 2 = 52=>tan3(θ)+cot3(θ)−2=52