If tan theta = frac{cos 25^{circ} + sin 25^{c | vedclass

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(D) Given $	an 	heta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$.Dividing the numerator and denominator by $\cos

Vedclass · Accepted Answer

$250$

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(D) Given $	an 	heta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$.Dividing the numerator and denominator by $\cos 25^{\circ}$,we get $	an 	heta = \frac{1 + 	an 25^{\circ}}{1 - 	an 25^{\circ}}$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = 45^{\circ}$ and $B = 25^{\circ}$,we have $	an 	heta = 	an(45^{\circ} + 25^{\circ}) = 	an 70^{\circ}$.Since $	heta$ is in the third quadrant,we use the property $	an(180^{\circ} + \alpha) = 	an \alpha$.Thus,$	an 	heta = 	an(180^{\circ} + 70^{\circ}) = 	an 250^{\circ}$.Therefore,$	heta = 250^{\circ}$.

If $\tan \theta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$ and $\theta$ is in the third quadrant,then $\theta =$ (in $^{\circ}$)

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