3 tan^6 10^{circ} - 27 tan^4 10^{circ} + 33 t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}$.Let $	heta = 10^{\circ}$,then $	an 30^{\circ} = \frac{1}{\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We know that $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}$.Let $	heta = 10^{\circ}$,then $	an 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{3	an 10^{\circ} - 	an^3 10^{\circ}}{1 - 3	an^2 10^{\circ}}$.Squaring both sides,we get $\frac{1}{3} = \frac{(3	an 10^{\circ} - 	an^3 10^{\circ})^2}{(1 - 3	an^2 10^{\circ})^2}$.$(1 - 3	an^2 10^{\circ})^2 = 3(9	an^2 10^{\circ} - 6	an^4 10^{\circ} + 	an^6 10^{\circ})$.$1 - 6	an^2 10^{\circ} + 9	an^4 10^{\circ} = 27	an^2 10^{\circ} - 18	an^4 10^{\circ} + 3	an^6 10^{\circ}$.Rearranging the terms: $3	an^6 10^{\circ} - 27	an^4 10^{\circ} + 33	an^2 10^{\circ} = 1$.

$3 \tan^6 10^{\circ} - 27 \tan^4 10^{\circ} + 33 \tan^2 10^{\circ} = $

Similar Questions

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

If $A$ is not an integral multiple of $\frac{\pi}{2}$,then $\operatorname{cosec} 2A + \cot 2A$ is equal to

If $L = \sin^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$ and $M = \cos^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$,then which of the following is correct?

If $\sinh x = \frac{5}{12}$,then $\cosh \frac{x}{2} =$

$\cos 15^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module