∏n=3∞ 1−tan4⁡π2n

∏n=3∞ 1−tan2⁡π2n22tan⁡π2n2⋅1+tan2⁡π2n1−tan2⁡π2n×4tan2⁡π2nπn=3∞4tan2⁡π2ntan2⁡π2n−1⋅1cos2⁡π2n−1πn=3∞4tan2⁡π2ntan2⁡π2n−1⋅2sin⁡π2n−1sin⁡π2n−2=limn→∞ πn=3∞8tan2⁡π2ntan2⁡π2n−1⋅sin⁡π2n−1sin⁡π2n−2=183limn→∞ tan⁡π2nπ2n2sin⁡π2n−1π2n−1×π3×2=π332

∏n=3∞ 1−tan4⁡π2n

∏n=3∞ 1−tan2⁡π2n22tan⁡π2n2⋅1+tan2⁡π2n1−tan2⁡π2n×4tan2⁡π2nπn=3∞4tan2⁡π2ntan2⁡π2n−1⋅1cos2⁡π2n−1πn=3∞4tan2⁡π2ntan2⁡π2n−1⋅2sin⁡π2n−1sin⁡π2n−2=limn→∞ πn=3∞8tan2⁡π2ntan2⁡π2n−1⋅sin⁡π2n−1sin⁡π2n−2=183limn→∞ tan⁡π2nπ2n2sin⁡π2n−1π2n−1×π3×2=π332

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

$\prod_{n = 3}^{\infty} (1 - \tan^{4} (\frac{π}{2^{n}}))$

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

$\frac{π^{3}}{32}$

$\frac{π^{3}}{64}$

$\frac{π^{3}}{16}$

$\frac{π^{3}}{8}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\prod_{n = 3}^{\infty} \frac{{(1 - \tan^{2} \frac{π}{2^{n}})}^{2}}{{(2 \tan \frac{π}{2^{n}})}^{2}} \cdot \frac{1 + \tan^{2} \frac{π}{2^{n}}}{1 - \tan^{2} \frac{π}{2^{n}}} \times 4 \tan^{2} \frac{π}{2^{n}}$
$π_{n = 3}^{\infty} \frac{4 \tan^{2} \frac{π}{2^{n}}}{\tan^{2} \frac{π}{2^{n - 1}}} \cdot \frac{1}{\cos^{2} \frac{π}{2^{n - 1}}}$
$π_{n = 3}^{\infty} \frac{4 \tan^{2} (\frac{π}{2^{n}})}{\tan^{2} \frac{π}{2^{n - 1}}} \cdot \frac{2 \sin \frac{π}{2^{n - 1}}}{\sin \frac{π}{2^{n - 2}}} = lim_{n \to \infty} π_{n = 3}^{\infty} \frac{8 \tan^{2} \frac{π}{2^{n}}}{\tan^{2} \frac{π}{2^{n - 1}}} \cdot \frac{\sin \frac{π}{2^{n - 1}}}{\sin \frac{π}{2^{n - 2}}}$
$= \frac{1}{8^{3}} lim_{n \to \infty} {(\frac{\tan (\frac{π}{2^{n}})}{\frac{π}{2^{n}}})}^{2} \frac{\sin (\frac{π}{2^{n - 1}})}{\frac{π}{2^{n - 1}}} \times π^{3} \times 2 = \frac{π^{3}}{32}$