Question

if tan alpha + cot alpha = 2 then find tan^20 alpha + cot^20 alpha

Answer 1

Answer :

Easy way :

$\tan( \alpha ) + \cot( \alpha ) = 2 \\ \\ \tan( \alpha ) + \frac{1}{ \tan( \alpha ) } = 2$

let

$\tan( \alpha ) = y \\ y + \frac{1}{y} = 2 \\ \\ {y}^{2} + 1 = 2y \\( y - 1) {}^{2} = 0 \\ \\ \tan( \alpha ) = 1 \: \: and \: \: \: \: \cot( \alpha ) = 1$

So

$\tan {}^{20} ( \alpha ) + \cot{}^{20} ( \alpha ) = 1 + 1 = 2$

Fun way :

$\tan( \alpha ) + \cot( \alpha ) = 2$

say it's equation 1

square both sides

$\tan {}^{2} ( \alpha ) + \cot{}^{2} ( \alpha ) + 2 = 4 \\ \\ \tan {}^{2} ( \alpha ) + \cot{}^{2} ( \alpha ) = 2$

it's equation 2 . Now multiply eq.1 and eq.2

$\{\tan {}^{} ( \alpha ) + \cot{}^{} ( \alpha ) \}(\tan {}^{2} ( \alpha ) + \cot{}^{2} ( \alpha )) = 4 \\ \\ \tan {}^{3} ( \alpha ) + \cot{}^{3} ( \alpha ) + \tan( \alpha ) + \cot( \alpha ) = 4 \\ \\ \tan {}^{3} ( \alpha ) + \cot{}^{3} ( \alpha ) + 2 = 4 \\ \\ \tan {}^{3} ( \alpha ) + \cot{}^{3} ( \alpha ) = 2$

Now either keep multiplying by equation 1 or you can easily proof by induction , that

$\tan {}^{n} ( \alpha ) + \cot{}^{n} ( \alpha ) = 2$

All you need to do is put n = 20

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ \tan \alpha + \cot \alpha = 2}$

TO DETERMINE

$\displaystyle \sf{ {\tan}^{20} \alpha + {\cot}^{20} \alpha }$

EVALUATION

Here it is given that

$\displaystyle \sf{ \tan \alpha + \cot \alpha = 2}$

$\displaystyle \sf{ \implies \: \tan \alpha + \frac{1}{\tan \alpha } = 2}$

$\displaystyle \sf{ \implies \: \frac{{\tan}^{2} \alpha + 1}{\tan \alpha } = 2}$

$\displaystyle \sf{ \implies \: {\tan}^{2} \alpha + 1 = 2\tan \alpha }$

$\displaystyle \sf{ \implies \: {\tan}^{2} \alpha + 1 - 2\tan \alpha = 0}$

$\displaystyle \sf{ \implies \: {(\tan \alpha - 1)}^{2} = 0}$

$\displaystyle \sf{ \implies \: (\tan \alpha - 1) = 0}$

$\displaystyle \sf{ \implies \: \tan \alpha = 1}$

$\displaystyle \sf{ \implies \cot \alpha = 1}$

This gives

$\displaystyle \sf{ {\tan}^{20} \alpha + {\cot}^{20} \alpha }$

$\displaystyle \sf{ = {1}^{20} + {1}^{20} }$

$\sf{ =1 + 1}$

$= 2$

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