If tan theta = frac{cos 15^{circ} + sin 15^{c | vedclass

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(B) Divide the numerator and denominator by $\cos 15^{\circ}$:$	an 	heta = \frac{1 + 	an 15^{\circ}}{1 - 	an 15^{\circ}}$Since $	an 45^{\circ} =

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$\frac{\pi}{3}$

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(B) Divide the numerator and denominator by $\cos 15^{\circ}$:$	an 	heta = \frac{1 + 	an 15^{\circ}}{1 - 	an 15^{\circ}}$Since $	an 45^{\circ} = 1$,we can write this as:$	an 	heta = \frac{	an 45^{\circ} + 	an 15^{\circ}}{1 - 	an 45^{\circ} 	an 15^{\circ}}$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$	an 	heta = 	an(45^{\circ} + 15^{\circ})$$	an 	heta = 	an 60^{\circ}$Since $60^{\circ} = \frac{\pi}{3}$ radians,we have:$	heta = \frac{\pi}{3}$

If $\tan \theta = \frac{\cos 15^{\circ} + \sin 15^{\circ}}{\cos 15^{\circ} - \sin 15^{\circ}},$ then $\theta =$

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