If the solution of the equation log _{cos x} | vedclass

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(D) Given equation: $\log _{\cos x} \cot x+4 \log _{\sin x} 	an x=1$Let $u = \ln \sin x$ and $v = \ln \cos x$. Then $\cot x = \frac{e^v}{e^u}$ and $\

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$4$

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(D) Given equation: $\log _{\cos x} \cot x+4 \log _{\sin x} 	an x=1$Let $u = \ln \sin x$ and $v = \ln \cos x$. Then $\cot x = \frac{e^v}{e^u}$ and $	an x = \frac{e^u}{e^v}$.The equation becomes: $\frac{v-u}{v} + 4\frac{u-v}{u} = 1$$1 - \frac{u}{v} + 4\frac{u}{v} - 4 = 1$$3\frac{u}{v} = 4$ $\Rightarrow 3u = 4v$ $\Rightarrow 3 \ln \sin x = 4 \ln \cos x$$\sin^3 x = \cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$$\sin^4 x + \sin^3 x - 2\sin^2 x - 1 = 0$. This approach is complex. Let's re-evaluate the original equation: $\frac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4 \frac{\ln \sin x - \ln \cos x}{\ln \sin x} = 1$Let $a = \ln \sin x$ and $b = \ln \cos x$. Then $\frac{b-a}{b} + 4\frac{a-b}{a} = 1$$1 - \frac{a}{b} + 4 - 4\frac{b}{a} = 1 \Rightarrow 4 = \frac{a}{b} + 4\frac{b}{a}$. Let $t = \frac{a}{b}$.$t + \frac{4}{t} = 4$ $\Rightarrow t^2 - 4t + 4 = 0$ $\Rightarrow (t-2)^2 = 0$ $\Rightarrow t = 2$.So,$\frac{\ln \sin x}{\ln \cos x} = 2$ $\Rightarrow \ln \sin x = 2 \ln \cos x$ $\Rightarrow \sin x = \cos^2 x = 1 - \sin^2 x$.$\sin^2 x + \sin x - 1 = 0$. Using the quadratic formula,$\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$ (since $\sin x > 0$ for $x \in (0, \pi/2)$).Comparing $\sin x = \frac{-1 + \sqrt{5}}{2}$ with $\sin x = \frac{\alpha + \sqrt{\beta}}{2}$,we get $\alpha = -1$ and $\beta = 5$.Therefore,$\alpha + \beta = -1 + 5 = 4$.

If the solution of the equation $\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1$,where $x \in \left(0, \frac{\pi}{2}\right)$,is $\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$,where $\alpha, \beta$ are integers,then $\alpha+\beta$ is equal to:

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