If the solution of the equation log cos x ( cot x ) + 4 log sin x ( tan x ) = 1 , x ( 0 , 2 ) , is sin - 1 (2 ) , where are integers, then is equal to [2023]

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Solution

Q.
If the solution of the equation $\log_{\cos x} (\cot x) + 4 \log_{\sin x} (\tan x) = 1,$ $x \in (0, \frac{π}{2}),$ is $\sin^{- 1} (\frac{α + \sqrt{β}}{2}),$ where $α, β$ are integers, then $α + β$ is equal to [2023]

1 6

2 4

3 5

4 3

Ans.
(2)

$We have \log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$

$\Rightarrow \log_{\cos x} \cos x - \log_{\cos x} \sin x + 4 [\log_{\sin x} \sin x - \log_{\sin x} \cos x] = 1$

$\Rightarrow 1 - \log_{\cos x} \sin x + 4 - 4 \log_{\sin x} \cos x = 1$

$\Rightarrow - \log_{\cos x} \sin x + 4 - 4 \log_{\sin x} \cos x = 0$

Put $\log_{\cos x} \sin x = y$ $∴$ $- y + 4 - \frac{4}{y} = 0$

$\Rightarrow y^{2} - 4 y + 4 = 0 \Rightarrow {(y - 2)}^{2} = 0$ $∴$ $y = 2 = \log_{\cos x} \sin x$

$\Rightarrow \cos^{2} x = \sin x \Rightarrow 1 - \sin^{2} x = \sin x \Rightarrow \sin^{2} x + \sin x - 1 = 0$

$∴$ $\sin x = \frac{- 1 \pm \sqrt{1 - 4 \times 1 \times (- 1)}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$

Since $x \in (0, π / 2)$ ,

$∴$ $\sin x = \frac{- 1 + \sqrt{5}}{2} \Rightarrow x = \sin^{- 1} (\frac{- 1 + \sqrt{5}}{2})$

$∴$ $α = - 1, β = 5,$ So, $α + β = 4$

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