The numerical value of tan⁡π3+2tan⁡2π3+4tan⁡4π3+8tan⁡8π3 is equal to

tan⁡π3+2tan⁡2π3+4tan⁡4π3+8tan⁡8π3=tan⁡π3+2tan⁡π−π3+4tan⁡π+π3+8tan⁡3π−π3=tan⁡π3−2tan⁡π3+4tan⁡π3−8tan⁡π3=−5tan⁡π3=−53

The numerical value of tan⁡π3+2tan⁡2π3+4tan⁡4π3+8tan⁡8π3 is equal to

tan⁡π3+2tan⁡2π3+4tan⁡4π3+8tan⁡8π3=tan⁡π3+2tan⁡π−π3+4tan⁡π+π3+8tan⁡3π−π3=tan⁡π3−2tan⁡π3+4tan⁡π3−8tan⁡π3=−5tan⁡π3=−53

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The numerical value of $\tan \frac{π}{3} + 2 \tan \frac{2 π}{3} + 4 \tan \frac{4 π}{3} + 8 \tan \frac{8 π}{3}$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- 5 \sqrt{3}$

$- \frac{5}{\sqrt{3}}$

$\frac{5}{\sqrt{3}}$

$5 \sqrt{3}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{matrix} \tan \frac{π}{3} + 2 \tan \frac{2 π}{3} + 4 \tan \frac{4 π}{3} + 8 \tan \frac{8 π}{3} \\ = \tan \frac{π}{3} + 2 \tan (π - \frac{π}{3}) + 4 \tan (π + \frac{π}{3}) + 8 \tan (3 π - \frac{π}{3}) \\ = \tan \frac{π}{3} - 2 \tan \frac{π}{3} + 4 \tan \frac{π}{3} - 8 \tan \frac{π}{3} \\ = - 5 \tan \frac{π}{3} = - 5 \sqrt{3} \end{matrix}$