If sqrt{2} tan 2 theta = sqrt{6} and 0^{circ} | vedclass

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(B) Given: $\sqrt{2} 	an 2 	heta = \sqrt{6}$Divide both sides by $\sqrt{2}$:$	an 2 	heta = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$Since $	an 60^{\c

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$\frac{4}{3}$

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(B) Given: $\sqrt{2} 	an 2 	heta = \sqrt{6}$Divide both sides by $\sqrt{2}$:$	an 2 	heta = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$Since $	an 60^{\circ} = \sqrt{3}$,we have:$2 	heta = 60^{\circ}$$	heta = 30^{\circ}$Now,substitute $	heta = 30^{\circ}$ into the expression $\sin 	heta + \sqrt{3} \cos 	heta - 2 	an^{2} 	heta$:$= \sin 30^{\circ} + \sqrt{3} \cos 30^{\circ} - 2 	an^{2} 30^{\circ}$Using standard trigonometric values $\sin 30^{\circ} = \frac{1}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,and $	an 30^{\circ} = \frac{1}{\sqrt{3}}$:$= \frac{1}{2} + \sqrt{3} \left( \frac{\sqrt{3}}{2} ight) - 2 \left( \frac{1}{\sqrt{3}} ight)^{2}$$= \frac{1}{2} + \frac{3}{2} - 2 \left( \frac{1}{3} ight)$$= \frac{4}{2} - \frac{2}{3}$$= 2 - \frac{2}{3} = \frac{6-2}{3} = \frac{4}{3}$

If $\sqrt{2} \tan 2 \theta = \sqrt{6}$ and $0^{\circ} < \theta < 45^{\circ},$ then the value of $\sin \theta + \sqrt{3} \cos \theta - 2 \tan^{2} \theta$ is

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