The number of solutions of the equation tan t | vedclass

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(B) Given trigonometric equation is,$	an 	heta + \sec 	heta = 2 \cos 	heta$. Multiplying by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $\frac{\si

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given trigonometric equation is,$	an 	heta + \sec 	heta = 2 \cos 	heta$. Multiplying by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $\frac{\sin 	heta}{\cos 	heta} + \frac{1}{\cos 	heta} = 2 \cos 	heta$ $\Rightarrow 1 + \sin 	heta = 2 \cos^2 	heta$ Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$: $1 + \sin 	heta = 2(1 - \sin^2 	heta)$ $2 \sin^2 	heta + \sin 	heta - 1 = 0$ Factoring the quadratic: $(2 \sin 	heta - 1)(\sin 	heta + 1) = 0$ This gives $\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -1$. For $\sin 	heta = \frac{1}{2}$ in $(0, 2 \pi)$,$	heta = \frac{\pi}{6}$ or $	heta = \frac{5 \pi}{6}$. For $\sin 	heta = -1$ in $(0, 2 \pi)$,$	heta = \frac{3 \pi}{2}$. However,the original equation involves $	an 	heta$ and $\sec 	heta$,which are undefined at $	heta = \frac{3 \pi}{2}$ (since $\cos \frac{3 \pi}{2} = 0$). Thus,the valid solutions are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5 \pi}{6}$. Therefore,the number of solutions is $2$.

The number of solutions of the equation $\tan \theta + \sec \theta = 2 \cos \theta$ where $\cos \theta \neq 0$ lying in the interval $(0, 2 \pi)$ is

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