Finding the Smallest Value of p in sec x cosec x = p

We are given the trigonometric equation $\sec x \cdot \cosec x = p$, where $p$ is a positive real number. We need to find the smallest value of $p$ for which this equation has a solution for $x$.

Transforming the Trigonometric Equation

Let's rewrite the equation using fundamental trigonometric identities. We know that:

• $\sec x = \frac{1}{\cos x}$
• $\cosec x = \frac{1}{\sin x}$

Substituting these into the given equation:

$\frac{1}{\cos x} \cdot \frac{1}{\sin x} = p$

This simplifies to:

$\frac{1}{\sin x \cos x} = p$

We can use the double angle identity for sine, which is $\sin(2x) = 2 \sin x \cos x$. From this, we can express the product $\sin x \cos x$ as:

$\sin x \cos x = \frac{\sin(2x)}{2}$

Substitute this expression back into the equation:

$\frac{1}{\frac{\sin(2x)}{2}} = p$

$ \frac{2}{\sin(2x)} = p $

Rearranging this equation to isolate $\sin(2x)$:

$\sin(2x) = \frac{2}{p}$

Condition for Existence of a Solution

For the equation $\sin(2x) = \frac{2}{p}$ to have a solution for $x$, the value on the right side, $\frac{2}{p}$, must be within the range of the sine function. The range of $\sin \theta$ for any real $\theta$ is $[-1, 1]$.

Therefore, we must have:

$ -1 \le \sin(2x) \le 1 $

Substituting $\sin(2x) = \frac{2}{p}$:

$ -1 \le \frac{2}{p} \le 1 $

Solving the Inequality for p

We are given that $p$ is a positive real number, which means $p > 0$. We need to solve the inequality $-1 \le \frac{2}{p} \le 1$ for $p$, keeping in mind $p > 0$.

We can split the compound inequality into two separate inequalities:

1. $ \frac{2}{p} \le 1 $

2. $ -1 \le \frac{2}{p} $

Let's solve the first inequality, $\frac{2}{p} \le 1$. Since $p > 0$, we can multiply both sides by $p$ without changing the direction of the inequality sign:

$ 2 \le p $

Now, let's solve the second inequality, $-1 \le \frac{2}{p}$. Since $p > 0$, we can multiply both sides by $p$:

$ -p \le 2 $

Multiplying by $-1$ and reversing the inequality sign:

$ p \ge -2 $

We need $p$ to satisfy both conditions $p \ge 2$ and $p \ge -2$, and also the initial condition $p > 0$.

• The condition $p > 0$ is given.
• If $p \ge 2$, then $p$ is definitely positive, so $p > 0$ is satisfied.
• If $p \ge 2$, then $p$ is definitely greater than or equal to $-2$, so $p \ge -2$ is satisfied.

Thus, the condition for the equation to have a solution, given $p$ is positive, is $p \ge 2$.

Determining the Smallest Value of p

The inequality $p \ge 2$ tells us that $p$ can take any value that is greater than or equal to 2. We are looking for the smallest such positive value of $p$. The smallest value in the range $[2, \infty)$ is 2.

So, the smallest value of $p$ for which the equation $\sec x \cdot \cosec x = p$ has a solution is 2.

Checking the Options

Let's review the given options:

• Option 1: $ \frac{1}{2} $. Is $ \frac{1}{2} \ge 2$? No.
• Option 2: $ 1 $. Is $ 1 \ge 2$? No.
• Option 3: $ 2 $. Is $ 2 \ge 2$? Yes. This is the minimum value.
• Option 4: Minimum does not exist. This is incorrect, as we found the minimum value is 2.

The smallest positive real number $p$ for which the equation has a solution is indeed 2.

Revision Table: Key Concepts for sec x cosec x = p

Understanding the concepts used in solving this problem is crucial for similar questions.

Additional Information: Domain Considerations for sec x and cosec x

While solving for $p$, we focused on the range of $\sin(2x)$. It's also worth noting the domain restrictions for $\sec x$ and $\cosec x$ in the original equation $\sec x \cdot \cosec x = p$.

• $\sec x = \frac{1}{\cos x}$ is undefined when $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$.
• $\cosec x = \frac{1}{\sin x}$ is undefined when $\sin x = 0$, which occurs at $x = n\pi$ for any integer $n$.

So, for the original equation to be defined, $x$ cannot be a multiple of $ \frac{\pi}{2} $. In other words, $x \ne \frac{k\pi}{2}$ for any integer $k$.

When we transform the equation to $\sin(2x) = \frac{2}{p}$, the domain of $\sin(2x)$ is all real numbers for $2x$, implying $x$ can be any real number. However, the transformation from $\frac{1}{\sin x \cos x}$ to $\frac{2}{\sin(2x)}$ is valid only when $\sin x \cos x \ne 0$, which means $x$ is not a multiple of $\frac{\pi}{2}$.

The condition $\sin(2x) = \frac{2}{p}$ requires $\sin(2x) \ne 0$ because $p$ is positive, meaning $\frac{2}{p}$ cannot be zero. $\sin(2x)=0$ occurs when $2x = k\pi$, or $x = \frac{k\pi}{2}$ for some integer $k$. This aligns perfectly with the original domain restrictions where $\sec x$ or $\cosec x$ would be undefined. If $\sin(2x) = \frac{2}{p}$ has a solution where $\frac{2}{p} \in [-1, 1]$ and $\frac{2}{p} \ne 0$ (since $p$ is positive), then $\sin(2x)$ is non-zero, meaning $2x$ is not a multiple of $\pi$, and thus $x$ is not a multiple of $\frac{\pi}{2}$. This ensures that the solution $x$ obtained for $\sin(2x) = \frac{2}{p}$ will also be a valid $x$ for the original equation $\sec x \cdot \cosec x = p$, provided $\frac{2}{p}$ is in the range $[-1, 1]$ and non-zero.

Since $p$ is positive, $\frac{2}{p}$ is positive. The condition becomes $0 < \frac{2}{p} \le 1$. This implies $p > 0$ and $2 \le p$. Combining these, we get $p \ge 2$. So, the smallest value of $p$ remains 2.