If alpha,-frac{pi}{2}

Question

(C) Given equation: $4 \cos 	heta + 5 \sin 	heta = 1$. Divide by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $4 + 5 	an 	heta = \sec 	heta$. Squa

Vedclass · Accepted Answer

$\frac{\sqrt{10} - 10}{12}$

Vedclass · Answer

(C) Given equation: $4 \cos 	heta + 5 \sin 	heta = 1$. Divide by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $4 + 5 	an 	heta = \sec 	heta$. Squaring both sides: $(4 + 5 	an 	heta)^2 = \sec^2 	heta = 1 + 	an^2 	heta$. $16 + 25 	an^2 	heta + 40 	an 	heta = 1 + 	an^2 	heta$. $24 	an^2 	heta + 40 	an 	heta + 15 = 0$. Using the quadratic formula for $	an 	heta$: $	an 	heta = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)} = \frac{-40 \pm \sqrt{1600 - 1440}}{48} = \frac{-40 \pm \sqrt{160}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$. Since $-\frac{\pi}{2}  0$. From $4 \cos \alpha + 5 \sin \alpha = 1$,we have $5 \sin \alpha = 1 - 4 \cos \alpha$. For $\alpha$ to be a solution,$	an \alpha = \frac{-10 + \sqrt{10}}{12}$ satisfies the original equation. Thus,the correct option is $C$.

If $\alpha$,$-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ is the solution of $4 \cos \theta + 5 \sin \theta = 1$,then the value of $\tan \alpha$ is

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