If the solution of the equation log_cos x cot x + 4 log_sin x tan x = 1, x ∈ (0, π/2) is sin⁻¹((α +

Q: Wait, why do we reject the negative root in the quadratic equation?

Great catch! We are given the domain $x \in (0, \frac{\pi}{2})$. In this interval (the first quadrant), $\sin x$ must be strictly positive. The quadratic formula gives us $\sin x = \frac{-1 \pm \sqrt{5}}{2}$. Since $\sqrt{5} > 1$, the root $\frac{-1 - \sqrt{5}}{2}$ is negative, which is impossible for $\sin x$. Thus, we must reject it.

Q: Why can't we just use log properties directly without changing the base?

Logarithm properties like $\log_b(xy) = \log_b x + \log_b y$ only work when the bases are the same. Here, the bases are $\cos x$ and $\sin x$, which are different. Changing the base to a common one (like $\ln$) is the standard, most reliable way to unify the expression.

Q: Is there a simpler way to solve $\sin x = \cos^2 x$ without converting to a quadratic?

Not really. Converting to a quadratic in $\sin x$ is the most direct path. By using the identity $\cos^2 x = 1 - \sin^2 x$, you transform a transcendental equation into a simple algebraic quadratic, which is exactly what we want to solve.

Question

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

Options:

(A) $3$

(B) $5$

(C) $6$

(D) $4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is:
$$\log_{\cos x} \cot x + 4 \log_{\sin x} 	an x = 1$$

At first glance, the different bases $\cos x$ and $\sin x$ might seem daunting. However, a change of perspective is the key to unlocking this problem.

### The Universal Translator

We apply the change of base formula, $\log_b a = \frac{\ln a}{\ln b}$, to translate the equation into a common logarithmic base:

$$\frac{\ln \cot x}{\ln \cos x} + 4 \frac{\ln 	an x}{\ln \sin x} = 1$$

Using the property $\ln(\frac{a}{b}) = \ln a - \ln b$, and noting that $\cot x = \frac{\cos x}{\sin x}$ and $	an x = \frac{\sin x}{\cos x}$, we rewrite the equation as:

$$\frac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4 \frac{\ln \sin x - \ln \cos x}{\ln \sin x} = 1$$

### Clearing the Fog

Let us simplify the fractions. The first term becomes $1 - \frac{\ln \sin x}{\ln \cos x}$, and the second term becomes $4(1 - \frac{\ln \cos x}{\ln \sin x})$.

Let $u = \ln \sin x$ and $v = \ln \cos x$. The equation transforms into:

$$1 - \frac{u}{v} + 4 - 4\frac{v}{u} = 1$$

The constants cancel out, leaving us with $4 - \frac{u}{v} - 4\frac{v}{u} = 0$. Multiplying the entire equation by $uv$ yields:

$$4uv - u^2 - 4v^2 = 0 \quad \Rightarrow \quad u^2 - 4uv + 4v^2 = 0$$

### The Perfect Square Reveal

The expression $u^2 - 4uv + 4v^2$ is a perfect square. It factors exactly into:

$$(u - 2v)^2 = 0$$

This implies $u = 2v$. Substituting our original variables back, we get $\ln \sin x = 2 \ln \cos x$, which simplifies to:

$$\ln \sin x = \ln (\cos^2 x)$$

Dropping the logarithms, we have $\sin x = \cos^2 x$. Using the identity $\cos^2 x = 1 - \sin^2 x$, we arrive at the quadratic equation:

$$\sin^2 x + \sin x - 1 = 0$$

### The Final Stretch

Solving this quadratic equation using the quadratic formula, we obtain:

$$\sin x = \frac{-1 \pm \sqrt{5}}{2}$$

Given the domain $x \in (0, \frac{\pi}{2})$, $\sin x$ must be positive. Therefore, we reject the negative root and accept:

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

Comparing this to the form $\sin x = \frac{\alpha + \sqrt{\beta}}{2}$, we identify $\alpha = -1$ and $\beta = 5$. The final sum is:

$$\alpha + \beta = -1 + 5 = \mathbf{4}$$

Animated Solution for Mathematics - Trigonometry: If the solution of the equation $lo g_{c o s x} cot x + 4 lo g_{s i n x} tan x = 1, x \in (0, \frac{π}{2})$ is $sin^{- 1} (\frac{α + β}{2})$ , where $α, β$ are integers, then $α + β$ is equal to:

Select Answer:

Visualized Solution

Analyze the Equation

Apply Change of Base Formula

Expand Logarithms of Quotients

Simplify the Fractions

Introduce Variables $u$ and $v$

Form a Quadratic Equation

Solve for $u$ in terms of $v$

Convert Back to Trigonometry

Solve the Quadratic Equation for $sin x$

Filter the Solution Based on Domain

Find $α + β$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Universal Translator

Clearing the Fog

The Perfect Square Reveal

The Final Stretch

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The Sigma Insight: Trigonometric Ratios and Identities

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The Universal Translator

Clearing the Fog

The Perfect Square Reveal

The Final Stretch

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