Question : If $\cos \theta+\sec \theta=\sqrt{3}$, then the value of $\cos ^3 \theta+\sec ^3 \theta$ is:
Option 1: $0$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $\sqrt{3}$
Option 4: $2 \sqrt{3}$
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Correct Answer: $0$
Solution :
Given, $\cos \theta+\sec \theta=\sqrt{3}$
Cubing both sides, we get,
$⇒\cos^3\theta + \sec^3\theta+3\cdot\cos\theta\cdot\sec\theta(\cos \theta+\sec \theta)=3\sqrt3$
⇒ $\cos^3\theta+\sec^3\theta+3\times 1(\sqrt3)=3\sqrt3$ [$\because \cos\theta=\frac{1}{\sec\theta}$]
⇒ $\cos^3\theta+\sec^3\theta+3\sqrt3=3\sqrt3$
⇒ $\cos^3\theta+\sec^3\theta=0$
Hence, the correct answer is $0$.
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