Question

tan a· cot a =2 then tan20 a cot 20 a =

Answer 1

Answer:

tanθ+cotθ=2

Squaring both sides;

(tanθ+cotθ)

2

=2

2

⇒tan

2

θ+cot

2

θ+2tanθcotθ=4

⇒tan

2

θ+cot

2

θ+2=4

⇒tan

2

θ+cot

2

θ=2

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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