Question

If $tanx+cotx=2$, then $si{n}^{2n}x+co{s}^{2n}x=$

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

Answer 1

Answer: (D)

$\frac{1}{2^{n-1}}$

Given $\tan x+\cot x=2$

$⟹\tan x+\frac{1}{\tan x}=2$

$⟹{\tan }^{2}x-2\tan x+1=0$

$⟹(\tan x-1{)}^2=0$

$⟹\tan x=1⟹x=\frac{\pi }{4}$

${\sin }^{2n}x+{\cos }^{2n}x=(\frac{1}{\sqrt{2}}{)}^{2n}+(\frac{1}{\sqrt{2}}{)}^{2n}=\frac{1}{2^n}+\frac{1}{2^n}=\frac{2}{2^n}=\frac{1}{2^{n-1}}$