Question 66

If $$\cosec 3\theta = \sec(20^\circ + 2\theta), then \theta$$ is equal to:

Solution

$$\cosec 3\theta = \sec(20^\circ + 2\theta)$$

We know that $$\cosec (90^\circ-\theta)=\sec \theta$$

$$\Rightarrow \cosec 3\theta = \cosec(90^\circ - 20^\circ - 2\theta)$$

$$\Rightarrow \cosec 3\theta = \cosec(70^\circ - 2\theta)$$

From the above,

$$\Rightarrow 3\theta=70^\circ-2\theta$$

$$\Rightarrow 5\theta=70^\circ$$

$$\Rightarrow \theta=14^\circ$$

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