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

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(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^{\

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

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(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

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