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 $\boxed{4}$.