Question 57

If $$\sec 4\theta = \cosec (\theta + 20^\circ)$$, then $$\theta$$ is equal to:

Solution

$$\sec 4\theta = \cosec (\theta + 20^\circ)$$

$$\operatorname{cosec}\left(90-4\theta\ \right)=\operatorname{cosec}\left(\theta\ +20^{^{\circ\ }}\right)$$

$$90-4\theta\ =\theta\ +20^{^{\circ\ }}=5\theta\ =70^{\circ\ }$$

$$\theta\ =14^{\circ\ }$$

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