Question

refer to the attachment and answer correctly​

Answer 1

Answer:

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Here we are given:

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$\sin x + cosec \, x = 2 \\ \\ \\ \implies \sin x + \frac{1}{\sin x} = 2 \\ \\ \\ \implies \frac{\sin^2x+1}{\sin x}=2 \\ \\ \\ \implies \sin^2x+1 = 2\sin x \\ \\ \\ \implies \sin^2x - 2\sin x + 1 = 0 \\ \\ \\ \implies (\sin x - 1)^2 = 0 \\ \\ \\ \implies \sin x -1 =0 \\ \\ \\ \implies \boxed{\sin x =1} \\ \\ \\ \\ \implies \boxed{cosec \, x = \frac{1}{\sin x} = 1}$

Now, we have:

$\sin^n x + cosec^n \, x \\ \\ \\ = (1)^n + (1)^n \\ \\ \\ = 1 + 1 \\ \\ \\ = 2 \\ \\ \\ \\ \implies \boxed{\sin^n x + cosec^n \, x = 2}$

Hope it helps uhh

Answer 2

Answer:

Is equal to 2

Explanation: