Question

If for log_{cos x} (cot x) - 4log_{(sin x)} cot x = 1, x = sin^-1 \((\frac{\alpha+\sqrt{\beta}}{2})\). Find (α + β), given x ∈ \((0,\frac{\pi}{2})\)

Answer 1

Correct Answer: 4

The given equation is:

\[ \log \cos x \cdot \cot x + 4 \log \sin x \cdot \tan x = 1 \]

Simplify the Logarithms

Using the change of base formula:

\[ \log \cos x \cdot \cot x = \ln \cos x - \ln \sin x \]

Substitute these into the equation:

\[ \ln \cos x - \ln \sin x + 4 \cdot \left( \ln \sin x - \ln \cos x \right) = 1 \]

Simplify the Terms

Combine the terms:

\[ \frac{(\ln \cos x)^2 - (\ln \sin x)(\ln \cos x) + 4 \cdot (\ln \sin x)^2 - (\ln \cos x)(\ln \sin x)}{(\ln \cos x)(\ln \sin x)} = 1 \]

Factorize:

\[ (\ln \sin x)^2 - 4 (\ln \sin x)(\ln \cos x) + 4 (\ln \cos x)^2 = (\ln \cos x)(\ln \sin x) \]

Simplify further:

\[ \ln \sin x = 2 \ln \cos x \]

Relate Sine and Cosine

Exponentiate both sides:

\[ \sin^2 x = e^{2 \ln \cos x} = (\cos x)^2 \]

Thus:

\[ \sin^2 x + \sin x - 1 = 0 \]

Solve the Quadratic Equation

Solve the quadratic equation \( \sin^2 x + \sin x - 1 = 0 \) using the quadratic formula:

\[ \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2} \]

Since \( x \in [0, \frac{\pi}{2}] \), we take the positive root:

\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

Find \( \alpha + \beta \)

Comparing with \( \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2} \right) \), we identify:

\[ \alpha = -1, \quad \beta = 5 \]

Thus:

\[ \alpha + \beta = -1 + 5 = 4 \]

Conclusion: The value of \( \alpha + \beta \) is 4. Therefore, the correct answer is \( \boxed{4} \).