The solution of 3 tan (A - 15^{circ}) = tan ( | vedclass

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

Question

(A) Given: $3 	an (A - 15^{\circ}) = 	an (A + 15^{\circ})$$\Rightarrow 3 \frac{\sin (A - 15^{\circ})}{\cos (A - 15^{\circ})} = \frac{\sin (A + 15^{\

Vedclass · Accepted Answer

$n\pi + \frac{\pi}{4}$

Vedclass · Answer

(A) Given: $3 	an (A - 15^{\circ}) = 	an (A + 15^{\circ})$$\Rightarrow 3 \frac{\sin (A - 15^{\circ})}{\cos (A - 15^{\circ})} = \frac{\sin (A + 15^{\circ})}{\cos (A + 15^{\circ})}$$\Rightarrow 3 \sin (A - 15^{\circ}) \cos (A + 15^{\circ}) = \sin (A + 15^{\circ}) \cos (A - 15^{\circ})$Using $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$,we multiply both sides by $2$:$3 [\sin(2A) + \sin(-30^{\circ})] = \sin(2A) + \sin(30^{\circ})$$3 \sin(2A) - 3 \sin(30^{\circ}) = \sin(2A) + \sin(30^{\circ})$$2 \sin(2A) = 4 \sin(30^{\circ})$$2 \sin(2A) = 4 	imes \frac{1}{2} = 2$$\sin(2A) = 1$Since $\sin(	heta) = 1 \Rightarrow 	heta = 2n\pi + \frac{\pi}{2}$,we have:$2A = 2n\pi + \frac{\pi}{2}$$A = n\pi + \frac{\pi}{4}$

The solution of $3 \tan (A - 15^{\circ}) = \tan (A + 15^{\circ})$ is

Similar Questions

If $\cos x = |\sin x|$,then the general solution is

The equation $e^{\sin x} - e^{-\sin x} = 4$ has . . . . . . solutions.

If $5 \cos 2 \theta + 2 \cos^2 \frac{\theta}{2} + 1 = 0$, where $0 < \theta < \pi$, then the values of $\theta$ are:

The set of values of $x$ for which the expression $\frac{\tan 3x - \tan 2x}{1 + \tan 3x \tan 2x} = 1$ is

Number of solutions of the equation $\cos \theta + \cos 2\theta - \sqrt{3}(\sin \theta + \sin 2\theta) + 1 = 0$ lying in the interval $(0, 2\pi)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module