If tan theta - cot theta = 0 and theta is a p | vedclass

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(B) Given the equation $	an 	heta - \cot 	heta = 0$.This implies $	an 	heta = \cot 	heta$.Since $\cot 	heta = 	an(90^{\circ} - 	heta)$,we hav

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$3$

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(B) Given the equation $	an 	heta - \cot 	heta = 0$.This implies $	an 	heta = \cot 	heta$.Since $\cot 	heta = 	an(90^{\circ} - 	heta)$,we have $	an 	heta = 	an(90^{\circ} - 	heta)$.Equating the angles,$	heta = 90^{\circ} - 	heta$,which gives $2	heta = 90^{\circ}$,so $	heta = 45^{\circ}$.Now,substitute $	heta = 45^{\circ}$ into the expression:$\frac{	an(45^{\circ} + 15^{\circ})}{	an(45^{\circ} - 15^{\circ})} = \frac{	an 60^{\circ}}{	an 30^{\circ}}$.Since $	an 60^{\circ} = \sqrt{3}$ and $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,The expression becomes $\frac{\sqrt{3}}{1/\sqrt{3}} = \sqrt{3} 	imes \sqrt{3} = 3$.

If $\tan \theta - \cot \theta = 0$ and $\theta$ is a positive acute angle,then the value of $\frac{\tan (\theta + 15^{\circ})}{\tan (\theta - 15^{\circ})}$ is

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