If for x ∈ (0, π/2), log₁₀ sin x + log₁₀ cos x = -1 and log₁₀ (sin x + cos x) = 1/2 (log₁₀ n - 1), t

Q: Wait, why do we reject the negative root when we take the square root of both sides?

That is a fantastic observation! In the interval $x \in (0, \frac{\pi}{2})$, both $\sin x$ and $\cos x$ are positive. Therefore, their sum $\sin x + \cos x$ must also be positive. When we solve $\log_{10}(\sin x + \cos x) = \log_{10}(\sqrt{\frac{n}{10}})$, we are equating the arguments. Since the argument of a logarithm must be positive, we only consider the positive root, $\sqrt{\frac{n}{10}}$.

Q: Why can't we just solve for $x$ directly using the first equation?

You could try to solve $\sin x \cos x = 0.1$, which leads to $\frac{1}{2} \sin(2x) = 0.1$, or $\sin(2x) = 0.2$. While you could find $x$ using an inverse sine function, it would make the second equation incredibly messy to evaluate. In JEE Advanced problems, look for algebraic shortcuts—like using the identity $(\sin x + \cos x)^2$—to bypass finding the specific value of $x$ entirely.

Q: Is there a trick to handling the $\frac{1}{2}$ coefficient in the second equation?

Absolutely! The $\frac{1}{2}$ is a classic logarithmic trap. The best way to handle it is to use the power rule: $k \log_b a = \log_b (a^k)$. By moving the $\frac{1}{2}$ inside as an exponent, you transform it into a square root, which perfectly sets you up to square both sides later. Always look to clear coefficients before equating logarithmic arguments.

Question

If for $x \in \left(0, \frac{\pi}{2}\right)$, $\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)$, then the value of $n$ is equal to :

Options:

(A) 20

(B) 12

(C) 9

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given $x \in (0, \frac{\pi}{2})$. This domain is crucial as it ensures $\sin x$ and $\cos x$ are strictly positive, keeping our logarithmic arguments valid.

We begin with the equation:
$$\log_{10} \sin x + \log_{10} \cos x = -1$$

Using the product rule of logarithms, $\log_{b} a + \log_{b} c = \log_{b} (ac)$, we simplify the expression:
$$\log_{10} (\sin x \cos x) = -1$$

Converting this to exponential form, where $\log_{b} a = c \Rightarrow b^{c} = a$, we obtain:
$$\sin x \cos x = 10^{-1} = \frac{1}{10}$$

### The Bridge to the Second Equation

Now, consider the second equation:
$$\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)$$

We express $1$ as $\log_{10} 10$ and apply the quotient rule, $\log_{b} a - \log_{b} c = \log_{b} (\frac{a}{c})$:
$$\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10) = \frac{1}{2} \log_{10} \left( \frac{n}{10} ight)$$

Applying the power rule, $k \log_{b} a = \log_{b} (a^{k})$, the coefficient $\frac{1}{2}$ becomes an exponent:
$$\log_{10} (\sin x + \cos x) = \log_{10} \left( \sqrt{\frac{n}{10}} ight)$$

### The Grand Unification

Since the logarithmic function is one-to-one, we equate the arguments:
$$\sin x + \cos x = \sqrt{\frac{n}{10}}$$

To relate this to our known product $\sin x \cos x = \frac{1}{10}$, we square both sides:
$$(\sin x + \cos x)^{2} = \frac{n}{10}$$

Expanding the left side using the identity $(\sin x + \cos x)^{2} = \sin^{2} x + \cos^{2} x + 2 \sin x \cos x$ and noting that $\sin^{2} x + \cos^{2} x = 1$, we get:
$$1 + 2 \sin x \cos x = \frac{n}{10}$$

### Final Calculation

Substitute the previously determined value $\sin x \cos x = \frac{1}{10}$ into the equation:
$$1 + 2 \left( \frac{1}{10} ight) = \frac{n}{10}$$

This simplifies to:
$$1 + \frac{1}{5} = \frac{n}{10} \Rightarrow \frac{6}{5} = \frac{n}{10}$$

Solving for $n$:
$$n = \frac{6 	imes 10}{5} = 12$$

The final value is **$n = 12$**.

Animated Solution for Mathematics - Trigonometry: If for $x \in (0, \frac{π}{2})$ , $lo g_{10} sin x + lo g_{10} cos x = - 1$ and $lo g_{10} (sin x + cos x) = \frac{1}{2} (lo g_{10} n - 1)$ , then the value of $n$ is equal to :

Select Answer:

Visualized Solution

Understanding the Given Conditions

Applying the Product Rule of Logarithms

Finding the Product $sin x cos x$

Simplifying the Second Equation

Using Quotient and Power Rules

Removing Logarithms from Both Sides

Squaring Both Sides

Applying Trigonometric Identity

Substituting the Product Value

Solving for $n$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Bridge to the Second Equation

The Grand Unification

Final Calculation

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