If \operatorname{cosec}^2x-2=0, then the value of x(0 < x < 90^\circ) is:

Question 65

If $$\operatorname{cosec}^2x-2=0$$, then the value of $$x(0 < x < 90^\circ)$$ is:

Solution

Given, $$\operatorname{cosec}^2x-2=0$$

$$\Rightarrow$$ $$\operatorname{cosec}^2x=2$$

$$\Rightarrow$$ $$\operatorname{cosec}x=\sqrt{2}$$

$$\Rightarrow$$ $$\frac{1}{\sin x}=\sqrt{2}$$

$$\Rightarrow$$ $$\sin x=\frac{1}{\sqrt{2}}$$

Since $$0 < x < 90^\circ$$

$$\sin x=\sin45^{\circ\ }$$

$$\Rightarrow$$ $$x=45^{\circ\ }$$

Hence, the correct answer is Option D

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