If sin x + operatorname{cosec} x = 3,then the | vedclass

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(B) Given that $\sin x + \operatorname{cosec} x = 3$. Squaring both sides,we get: $(\sin x + \operatorname{cosec} x)^{2} = 3^{2}$ $\sin^{2} x + \opera

Vedclass · Accepted Answer

$47$

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(B) Given that $\sin x + \operatorname{cosec} x = 3$. Squaring both sides,we get: $(\sin x + \operatorname{cosec} x)^{2} = 3^{2}$ $\sin^{2} x + \operatorname{cosec}^{2} x + 2 \sin x \operatorname{cosec} x = 9$ Since $\sin x \operatorname{cosec} x = 1$,we have: $\sin^{2} x + \operatorname{cosec}^{2} x + 2(1) = 9$ $\sin^{2} x + \operatorname{cosec}^{2} x = 7$ Now,squaring both sides again: $(\sin^{2} x + \operatorname{cosec}^{2} x)^{2} = 7^{2}$ $\sin^{4} x + \operatorname{cosec}^{4} x + 2 \sin^{2} x \operatorname{cosec}^{2} x = 49$ $\sin^{4} x + \operatorname{cosec}^{4} x + 2(1)^{2} = 49$ $\sin^{4} x + \operatorname{cosec}^{4} x = 49 - 2 = 47$

If $\sin x + \operatorname{cosec} x = 3$,then the value of $\sin^{4} x + \operatorname{cosec}^{4} x$ is

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