The value of csc frac{pi}{18} - sqrt{3} sec f | vedclass

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(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.$E = \frac{\co

Vedclass · Accepted Answer

natural number

Vedclass · Answer

(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.$E = \frac{\cos \frac{\pi}{18} - \sqrt{3} \sin \frac{\pi}{18}}{\sin \frac{\pi}{18} \cos \frac{\pi}{18}}$.Multiply numerator and denominator by $2$:$E = \frac{2 \left( \frac{1}{2} \cos \frac{\pi}{18} - \frac{\sqrt{3}}{2} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{2\pi}{18}} = \frac{2 \left( \sin \frac{\pi}{6} \cos \frac{\pi}{18} - \cos \frac{\pi}{6} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{\pi}{9}}$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$E = \frac{2 \sin(\frac{\pi}{6} - \frac{\pi}{18})}{\frac{1}{2} \sin \frac{\pi}{9}} = \frac{4 \sin(\frac{3\pi - \pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{2\pi}{18}}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{\pi}{9}}{\sin \frac{\pi}{9}} = 4$.Since $4$ is a natural number,the correct option is $D$.

The value of $\csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18}$ is a

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