tan 3A - tan 2A - tan A = | vedclass

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

Question

(A) We know that $3A = 2A + A$.Taking tangent on both sides: $	an 3A = 	an(2A + A)$.Using the formula $	an(x + y) = \frac{	an x + 	an y}{1 - 	an

Vedclass · Accepted Answer

$	an 3A 	an 2A 	an A$

Vedclass · Answer

(A) We know that $3A = 2A + A$.Taking tangent on both sides: $	an 3A = 	an(2A + A)$.Using the formula $	an(x + y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get:$	an 3A = \frac{	an 2A + 	an A}{1 - 	an 2A 	an A}$.Cross-multiplying gives:$	an 3A(1 - 	an 2A 	an A) = 	an 2A + 	an A$.$	an 3A - 	an 3A 	an 2A 	an A = 	an 2A + 	an A$.Rearranging the terms,we get:$	an 3A - 	an 2A - 	an A = 	an 3A 	an 2A 	an A$.

$\tan 3A - \tan 2A - \tan A = $

Similar Questions

If $\tan x = (\sin 45^{\circ})(\cos 45^{\circ}) + \sin 30^{\circ}$,then the value of $x$ is (in $^{\circ}$)

If $\tan A = \frac{1}{2},$ then $\tan 3A = $

What is the value of $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$?

The value of $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$ is

The value of $(\operatorname{cosec} a - \sin a)(\sec a - \cos a)(\tan a + \cot a)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module