If cos P


If $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$, where $P$ and $Q$ both are acute angles. Then, the value of $P-Q$ is

(a) $\frac{\pi}{6}$

(b) $\frac{\pi}{3}$

(c) $\frac{\pi}{4}$

(d) $\frac{\pi}{12}$


(b) $60^{\circ}=\frac{\pi}{3}$

$\cos P=\frac{1}{7}, \quad \cos Q=\frac{13}{14}$

Therefore, $\sin P=\sqrt{1-\frac{1}{49}}=\frac{4 \sqrt{3}}{7}$ and $\sin Q=\sqrt{1-\frac{169}{196}}=\frac{3 \sqrt{3}}{14}$

Hence, $\tan P=4 \sqrt{3}, \quad \tan Q=\frac{3 \sqrt{3}}{13}$

$\cos (P-Q)=\cos P \cos Q+\sin P \sin Q$

$=\frac{1}{7} \times \frac{13}{14}+\frac{4 \sqrt{3}}{7} \times \frac{3 \sqrt{3}}{14}$



$\therefore \cos (P-Q)=\frac{1}{2}$

$\Rightarrow P-Q=\cos ^{-1} \frac{1}{2}$

$\Rightarrow P-Q=60^{\circ}$

Hence, the correct answer is option B.


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