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Solution
Since $$cosecθ = \frac{1}{sinθ}$$
$$sin θ + cosec θ = 2 $$ becomes
$$sin θ + \frac{1}{sinθ} =2$$
$$sin^2θ - 2sinθ +1 =0 $$
which is $$(sinθ - 1)^2 = 0$$
$$sin θ =1$$
$$sin^{5}\theta+cosec^{5}\theta = 1 + 1 = 2$$
Hence Option D is th correct answer.
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