If the equation in variable theta,3 tan(theta | vedclass

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(B) Given the equation: $3 	an(	heta - \alpha) = 	an(	heta + \alpha)$.This can be rewritten as $\frac{	an(	heta + \alpha)}{	an(	heta - \alpha)

Vedclass · Accepted Answer

$\frac{5\pi}{18}$

Vedclass · Answer

(B) Given the equation: $3 	an(	heta - \alpha) = 	an(	heta + \alpha)$.This can be rewritten as $\frac{	an(	heta + \alpha)}{	an(	heta - \alpha)} = 3$.Applying componendo and dividendo:$\frac{	an(	heta + \alpha) + 	an(	heta - \alpha)}{	an(	heta + \alpha) - 	an(	heta - \alpha)} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2$.Using the identity $	an A \pm 	an B = \frac{\sin(A \pm B)}{\cos A \cos B}$,we get:$\frac{\sin(2	heta)}{\sin(2\alpha)} = 2 \implies \sin(2	heta) = 2 \sin(2\alpha)$.For this equation to have no real solution for $	heta$,the value of $2 \sin(2\alpha)$ must be outside the range $[-1, 1]$.Thus,$|2 \sin(2\alpha)| > 1 \implies |\sin(2\alpha)| > \frac{1}{2}$.Checking the options:For $A: \alpha = \frac{\pi}{15}, |\sin(\frac{2\pi}{15})| \approx |\sin(24^\circ)| For $B: \alpha = \frac{5\pi}{18}, |\sin(\frac{5\pi}{9})| = |\sin(100^\circ)| \approx 0.98 > 0.5$.For $C: \alpha = \frac{5\pi}{12}, |\sin(\frac{5\pi}{6})| = |\sin(150^\circ)| = 0.5$ (not strictly greater).For $D: \alpha = \frac{17\pi}{18}, |\sin(\frac{17\pi}{9})| = |\sin(340^\circ)| \approx 0.34 Therefore,the correct option is $B$.

If the equation in variable $\theta$,$3 \tan(\theta - \alpha) = \tan(\theta + \alpha)$,(where $\alpha$ is a constant) has no real solution,then $\alpha$ can be (wherever $\tan(\theta - \alpha)$ and $\tan(\theta + \alpha)$ are both defined).

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