If =cosec x=2, then write the value of ^nx+ cosec^n x.

Question

If $\sin\text{x}=\text{cosec}\text{ x}=2,$ then write the value of $\sin^\text{n}\text{x}\text{+ cosec}^\text{n}\text{ x}.$

✓

Answer

$\sin\text{x}+\text{cosec x}=2$
$\Rightarrow\sin\text{x}+\frac{1}{\sin\text{x}}=2$
$\Rightarrow\frac{\sin\text{x}+1}{\sin\text{x}}=2$
$\Rightarrow\sin^2\text{x}+1=2\sin\text{x}$
$\Rightarrow\sin^2\text{x}+1-2\sin\text{x}=0$
$\Rightarrow(\sin\text{x}-1)^2=0$
$\Rightarrow\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=1$
And, $\text{cosec x}=\frac{1}{\sin\text{x}}=1$
$\therefore\sin^\text{n}\text{x}+\text{cosec}^\text{n}\text{x}=1^\text{n}+1^\text{n}$
$=1+1=2$

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