If $sin x + c o s e c x = 2,$ then ${sin}^{n} x + {c o s e c}^{n} x$ is equal to

2

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2^{n}

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2^{n - 1}

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2^{n - 2}

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Solution

The correct option is A $2$
We have $s i n x + c o s e c x = 2$
$=> s i n x + \frac{1}{s i n x} = 2$
This means $x = 90^{0}$ as maximum value of $s i n x = 1$ such that $1 + 1 = 2$
Now, for any $n$ , we have $s i n^{n} 90^{0} + c o s e c^{n} 90^{0} = 1^{2} + 1^{2} = 2$

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