If 7 cos theta - sin theta = 5 and tan theta | vedclass

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(B) Given $7 \cos 	heta - \sin 	heta = 5$. Rearranging the equation,we get $\sin 	heta = 7 \cos 	heta - 5$. Using the identity $\sin^2 	heta + \c

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$\frac{3}{4}$

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(B) Given $7 \cos 	heta - \sin 	heta = 5$. Rearranging the equation,we get $\sin 	heta = 7 \cos 	heta - 5$. Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we substitute $\sin 	heta$: $(7 \cos 	heta - 5)^2 + \cos^2 	heta = 1$. $49 \cos^2 	heta - 70 \cos 	heta + 25 + \cos^2 	heta = 1$. $50 \cos^2 	heta - 70 \cos 	heta + 24 = 0$. Dividing by $2$,we get $25 \cos^2 	heta - 35 \cos 	heta + 12 = 0$. Factoring the quadratic equation: $(5 \cos 	heta - 3)(5 \cos 	heta - 4) = 0$. So,$\cos 	heta = \frac{3}{5}$ or $\cos 	heta = \frac{4}{5}$. If $\cos 	heta = \frac{3}{5}$,then $\sin 	heta = 7(\frac{3}{5}) - 5 = \frac{21-25}{5} = -\frac{4}{5}$. Here $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = -\frac{4}{3} If $\cos 	heta = \frac{4}{5}$,then $\sin 	heta = 7(\frac{4}{5}) - 5 = \frac{28-25}{5} = \frac{3}{5}$. Here $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{3/5}{4/5} = \frac{3}{4} > 0$. Since $	an 	heta > 0$,the correct value is $	an 	heta = \frac{3}{4}$.

If $7 \cos \theta - \sin \theta = 5$ and $\tan \theta > 0$,then $\tan \theta =$

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