Question

If tan α + cot α = 2, then tan^20α + cot^20α =​

Answer 1

Answer:

ans is 2

Step-by-step explanation:

$tan \alpha + cot \alpha = 2 \\ tan \alpha + \frac{1}{tan \alpha } = 2 \\ {(tan \alpha )}^{2} + 1 = 2tan \alpha \\ { \sec( \alpha ) }^{2} = 2 \tan( \alpha ) \\ \frac{1}{ { \cos( \alpha ) }^{2} } = 2 \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \frac{1}{ \cos( \alpha ) } = 2 \sin( \alpha ) \\ 1 = 2 \sin( \alpha ) \cos( \alpha ) \\ 1 = \sin(2 \alpha ) \\ \sin(90) = \sin(2 \alpha ) \\ 90 = 2 \alpha \\ \alpha = 45 \\ \\ thus \: \: \: { \tan(45) }^{20} + { \cot(45) }^{20} \\ {1}^{20} + {1}^{20} = 1 + 1 = 2 \: \: required \: answer$

Answer 2

The other user has already given an easier solution to this problem. Let me answer the problem with another approach.

Given data:

$tan\alpha+cot\alpha=2$

To find:

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

Step-by-step explanation:

Given, $tan\alpha+cot\alpha=2$

Squaring both sides, we get

$\quad tan^{2}\alpha+cot^{2}\alpha+2\:tan\alpha\:cot\alpha=4$

$\Rightarrow tan^{2}\alpha+cot^{2}\alpha+2=4\quad[\because tan\alpha\:cot\alpha=1]$

$\Rightarrow tan^{2}\alpha+cot^{2}\alpha=2$

Similarly, we get $tan^{4}\alpha+cot^{4}\alpha=2$,

$tan^{8}\alpha+cot^{8}\alpha=2$ and

$tan^{16}\alpha+cot^{16}\alpha=2$

Now, $(tan^{16}\alpha+cot^{16}\alpha)(tan^{4}\alpha+cot^{4}\alpha)$

$=tan^{20}\alpha+cot^{20}\alpha+tan^{16}\alpha\:cot^{4}\alpha+tan^{4}\alpha\:cot^{16}\alpha$

$=tan^{20}\alpha+cot^{20}\alpha+tan^{12}\alpha+cot^{12}\alpha$

$\Rightarrow 2\times 2=tan^{20}\alpha+cot^{20}\alpha+tan^{12}\alpha+cot^{12}\alpha$

$\Rightarrow tan^{20}\alpha+cot^{20}\alpha=4-(tan^{12}\alpha+cot^{12}\alpha)$ ... ... (i)

Now $(tan^{8}\alpha+cot^{8}\alpha)(tan^{4}\alpha+cot^{4}\alpha)$

$=tan^{12}\alpha+cot^{12}\alpha+tan^{8}\alpha\:cot^{4}\alpha+tan^{4}\alpha\:cot^{8}\alpha$

$=tan^{12}\alpha+cot^{12}\alpha+tan^{4}\alpha+cot^{4}\alpha$

$=tan^{12}\alpha+cot^{12}\alpha+2\quad [\because tan^{4}\alpha+cot^{4}\alpha=2]$

$\Rightarrow 2\times 2=tan^{12}\alpha+cot^{12}\alpha+2$

$\Rightarrow tan^{12}\alpha+cot^{12}\alpha=4-2$

$\Rightarrow tan^{12}\alpha+cot^{12}\alpha=2$

From (i), we get

$\quad tan^{20}\alpha+cot^{20}\alpha=4-2$

$\Rightarrow tan^{20}\alpha+cot^{20}\alpha=2$

Answer: $\tan^{20}\alpha+cot^{20}\alpha=2$