Question

tan alpha + cot alpha =2 then tan ^20 alpha +cot ^20 alpha =​

Answer 1

Answer:

Solution

verified

Verified by Toppr

Given tan(20−3α)=cot(5α−20)

⟹tan(20−3α)=tan(90−5α+20)

⟹tan(20−3α)=tan(110−5α)

⟹20−3α=110−5α

⟹5α−3α=110−20

⟹2α=90

⟹α=

2

90

=45

sinαsecαtanα−cosecαcosαcotα

=sin45sec45tan45−cosec45cos45cot45

=(

2

1

)(

2

)(1)−(

2

)(

2

1

)(1)=1−1=0

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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