(A) Given $\sin x + \text{cosec } x = 2$.Since $\text{cosec } x = \frac{1}{\sin x}$,we have $\sin x + \frac{1}{\sin x} = 2$.Let $\sin x = t$,then $t +

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(A) Given $\sin x + \text{cosec } x = 2$.Since $\text{cosec } x = \frac{1}{\sin x}$,we have $\sin x + \frac{1}{\sin x} = 2$.Let $\sin x = t$,then $t + \frac{1}{t} = 2$,which implies $t^2 - 2t + 1 = 0$.This simplifies to $(t - 1)^2 = 0$,so $t = 1$.Thus,$\sin x = 1$.Therefore,$\sin^n x + \text{cosec}^n x = (1)^n + (\frac{1}{1})^n = 1 + 1 = 2$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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If $\sin x + \text{cosec } x = 2$,then $\sin^n x + \text{cosec}^n x$ is equal to

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

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If $\sin x + \text{cosec } x = 2$,then $\sin^n x + \text{cosec}^n x$ is equal to

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