Question : If $\sin\theta+\operatorname{cosec}\theta=2$, the value of $\sin^{n}\theta+\operatorname{cosec}^{n}\theta$ is:
Option 1: $2^{n}$
Option 2: $2^{\frac{1}{n}}$
Option 3: $2$
Option 4: $0$
Correct Answer: $2$
Solution :
Given: $\sin\theta+\operatorname{cosec}\theta=2$
Squaring both sides, we get:
$(\sin\theta+cosec\theta)^{2}=2^{2}$
⇒ $\sin^{2}\theta+\operatorname{cosec}^{2}\theta+2\sin\theta×\operatorname{cosec}\theta=4$
⇒ $\sin^{2}\theta+\operatorname{cosec}^{2}\theta=4–2$
⇒ $\sin^{2}\theta+\operatorname{cosec}^{2}\theta=2$
Similarly, $(\sin\theta+\operatorname{cosec}\theta)^{3}=2^{3}$
⇒ $\sin^{3}\theta+\operatorname{cosec}^{3}\theta+\sin\theta×\operatorname{cosec}\theta×(\sin\theta+\operatorname{cosec}\theta)=8$
⇒ $\sin^{3}\theta+\operatorname{cosec}^{3}\theta+3×2=8$
⇒ $\sin^{3}\theta+\operatorname{cosec}^{3}\theta=2$
Similarly, we can say the value of $\sin^{n}\theta+\operatorname{cosec}^{n}\theta$ will be 2.
Hence, the correct answer is $2$.
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