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 $\cot 	heta - 	an 	heta = 2\cot 2	heta$,which implies $	an 	heta = \cot 	heta - 2\cot 2	heta$.Applying this to the las

Vedclass · Accepted Answer

$\cot \alpha$

Vedclass · Answer

(C) We use the identity $\cot 	heta - 	an 	heta = 2\cot 2	heta$,which implies $	an 	heta = \cot 	heta - 2\cot 2	heta$.Applying this to the last term: $8\cot 8\alpha = 4(2\cot 8\alpha) = 4(\cot 4\alpha - 	an 4\alpha)$.Substituting this into the expression:$= 	an \alpha + 2	an 2\alpha + 4	an 4\alpha + 4(\cot 4\alpha - 	an 4\alpha)$$= 	an \alpha + 2	an 2\alpha + 4\cot 4\alpha$Now,apply the identity again for $4\cot 4\alpha = 2(2\cot 4\alpha) = 2(\cot 2\alpha - 	an 2\alpha)$:$= 	an \alpha + 2	an 2\alpha + 2(\cot 2\alpha - 	an 2\alpha)$$= 	an \alpha + 2\cot 2\alpha$Finally,apply the identity for $2\cot 2\alpha = \cot \alpha - 	an \alpha$:$= 	an \alpha + (\cot \alpha - 	an \alpha)$$= \cot \alpha$.

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

Similar Questions

If $\cos A = \cos B \cos C$ and $A + B + C = \pi,$ then the value of $\cot B \cot C$ is

For any real values of $\theta$,$\sqrt{\frac{\sec \theta-1}{\sec \theta+1}} = ?$

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

The value of $k$,for which $(\cos x + \sin x)^2 + k \sin x \cos x - 1 = 0$ is an identity,is

The value of $\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module