tan alpha + 2tan 2alpha + 4tan 4alpha + 8cot | vedclass

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

Question

(C) We use the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,which implies $\cot 	heta - 	an 	heta = 2\cot 2	heta$.Consider the expression

Vedclass · Accepted Answer

$\cot \alpha $

Vedclass · Answer

(C) We use the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,which implies $\cot 	heta - 	an 	heta = 2\cot 2	heta$.Consider the expression $E = 	an \alpha + 2	an 2\alpha + 4	an 4\alpha + 8\cot 8\alpha$.Using the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,we have:$8\cot 8\alpha + 4	an 4\alpha = 4(2\cot 8\alpha + 	an 4\alpha) = 4(\cot 4\alpha) = 4\cot 4\alpha$.Now the expression becomes:$E = 	an \alpha + 2	an 2\alpha + 4\cot 4\alpha$.Again,using $4\cot 4\alpha + 2	an 2\alpha = 2(2\cot 4\alpha + 	an 2\alpha) = 2(\cot 2\alpha) = 2\cot 2\alpha$.Now the expression becomes:$E = 	an \alpha + 2\cot 2\alpha$.Using $2\cot 2\alpha = \cot \alpha - 	an \alpha$,we get:$E = 	an \alpha + (\cot \alpha - 	an \alpha) = \cot \alpha$.Thus,the correct option is $C$.

$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot 8\alpha = $

Similar Questions

$2\cos x - \cos 3x - \cos 5x = $

The simplest form of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4x}}}}$ is

$\frac{\sec 8\theta - 1}{\sec 4\theta - 1}$ is equal to

If $\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot \left(\frac{\pi}{24}\right)\right)=\sqrt{a^2+a}+\sqrt{a}$,then $a=$

The value of $\cot\, x + \cot\, (60^\circ + x) + \cot\, (120^\circ + x)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module