Question

If $\sin x+\cos ecx=2$ then ${\sin }^{n}x+\cos e{c}^{n}x$ is equal to

• A. $2$
• B. $2^n$
• C. $2^{n–1}$
• D. $2^{n–2}$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Given information,

$\Rightarrow \sin x+\cos ecx=2$

$\Rightarrow \sin x+\frac{1}{\sin x}=2(\because \cos ecx=\frac{1}{\sin x})$

$\Rightarrow \frac{{\sin }^{2}x+1}{\sin x}=2$

$\Rightarrow {\sin }^{2}x+1=2\sin x$

$\Rightarrow {\sin }^{2}x-2\sin x+1=0$

$\Rightarrow {\sin }^{2}x-2\sin x+1=0$

$\Rightarrow {(\sin x-1)}^2=0(\because {\sin }^{2}x+1^2-2\sin x={(\sin x-1)}^2)$

$\Rightarrow (\sin x-1)=0$(get square root on both sides)

$\Rightarrow \sin x=1$

$\Rightarrow \cos ecx=1(\because \cos ecx=\frac{1}{\sin x})$

From given question

$\Rightarrow {\sin }^{n}x+\cos e{c}^{n}x={\left(1\right)}^2+{\left(1\right)}^2$

$\Rightarrow 2$

Hence option ‘(a)’ is correct.