If tanx + cotx = 2 , then sin2n⁡x+cos2n⁡x is equal to

A
2ⁿ
B
$- \frac{1}{2}$
C
$\frac{1}{2}$
D
$\frac{1}{2^{n - 1}}$

Solution:

Since,

$\begin{matrix} \tan x + \frac{1}{\tan x} = 2 \Rightarrow \tan x = 1 \Rightarrow x = \frac{π}{4} \\ \sin x = \frac{1}{\sqrt{2}} and \cos x = \frac{1}{\sqrt{2}} \\ Hence, \sin^{2 n} x + \cos^{2 n} x = \frac{1}{2^{n}} + \frac{1}{2^{n}} = \frac{1}{2^{n - 1}} \end{matrix}$

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If tanx + cotx = 2 , then $\sin^{2 n} x + \cos^{2 n} x$ is equal to

Solution:

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