Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Using the identity $\tan A\tan B=\frac{\tan A-\tan B}{\tan (A-B)}-1$, each term $\tan (k\theta )\tan ((k+1)\theta )$ can be expressed as $\frac{\tan ((k+1)\theta )-\tan (k\theta )}{\tan \theta }-1$. The sum then telescopes to $\frac{1}{\tan \theta }(\tan (11\theta )-\tan \theta )-10$. Given $\theta =\frac{\pi }{12}$, we have $\tan \theta =2-\sqrt{3}$ and $\tan (11\theta )=\tan (\pi -\frac{\pi }{12})=-\tan (\frac{\pi }{12})=\sqrt{3}-2$. Substituting these values, the sum becomes $\frac{1}{2-\sqrt{3}}((\sqrt{3}-2)-(2-\sqrt{3}))-10=(2+\sqrt{3})(2\sqrt{3}-4)-10=(4\sqrt{3}-8+6-4\sqrt{3})-10=-2-10=-12$. The required absolute value is 12.