Question

The number of solution(s) of the equation $\text{cosec}^{-1}\left(\frac{x^2+1}{x^2}\right) + \text{sec}^{-1}\left(\frac{x^2+1}{x^2}-1\right) = \pi$

Answer 1

0

Answer 2

To find the number of solutions for the given equation, we first analyze the domain and range of the inverse trigonometric functions involved.

The given equation is: $\text{cosec}^{-1}\left(\frac{x^2+1}{x^2}\right) + \text{sec}^{-1}\left(\frac{x^2+1}{x^2}-1\right) = \pi$

Let's analyze the arguments of the inverse functions:

1. Argument for $\text{cosec}^{-1}$: Let $A = \frac{x^2+1}{x^2} = 1 + \frac{1}{x^2}$. Since $x^2$ is always positive for real $x \ne 0$, we have $\frac{1}{x^2} > 0$. Therefore, $A = 1 + \frac{1}{x^2} > 1$. The domain of $\text{cosec}^{-1}(y)$ is $|y| \ge 1$. Since $A > 1$, $\text{cosec}^{-1}(A)$ is well-defined for all $x \ne 0$. The principal range of $\text{cosec}^{-1}(y)$ is $[-\pi/2, \pi/2] \setminus \{0\}$. Since $A > 1$, $\text{cosec}^{-1}(A)$ lies in the interval $(0, \pi/2)$. (Note: $\text{cosec}^{-1}(1) = \pi/2$, but $A > 1$, so $\text{cosec}^{-1}(A) < \pi/2$). Let $P = \text{cosec}^{-1}\left(\frac{x^2+1}{x^2}\right)$. So, $P \in (0, \pi/2)$.

2. Argument for $\text{sec}^{-1}$: Let $B = \frac{x^2+1}{x^2}-1 = \left(1 + \frac{1}{x^2}\right) - 1 = \frac{1}{x^2}$. The domain of $\text{sec}^{-1}(y)$ is $|y| \ge 1$. So, we must have $\left|\frac{1}{x^2}\right| \ge 1$. Since $x^2 > 0$, this simplifies to $\frac{1}{x^2} \ge 1$. This inequality implies $1 \ge x^2$. Combining with the condition $x^2 > 0$ (from the first term), the overall domain for $x$ is $0 < x^2 \le 1$. This means $x \in [-1, 0) \cup (0, 1]$. For these values of $x$, $B = \frac{1}{x^2} \ge 1$. The principal range of $\text{sec}^{-1}(y)$ is $[0, \pi] \setminus \{\pi/2\}$. Since $B = \frac{1}{x^2} \ge 1$, $\text{sec}^{-1}(B)$ lies in the interval $[0, \pi/2)$. (Note: $\text{sec}^{-1}(1) = 0$, which occurs when $x^2=1$. As $x^2 \to 0$, $B \to \infty$, and $\text{sec}^{-1}(B) \to \pi/2$. So $\pi/2$ is not included). Let $Q = \text{sec}^{-1}\left(\frac{1}{x^2}\right)$. So, $Q \in [0, \pi/2)$.

Now, the given equation is $P + Q = \pi$. We have $P \in (0, \pi/2)$ and $Q \in [0, \pi/2)$. Let's find the range of $P+Q$: The minimum possible value of $P+Q$ is strictly greater than $0+0=0$. The maximum possible value of $P+Q$ is strictly less than $\pi/2 + \pi/2 = \pi$. So, the sum $P+Q$ must lie in the interval $(0, \pi)$.

Since the maximum possible value of $P+Q$ is strictly less than $\pi$, it is impossible for $P+Q$ to be equal to $\pi$. Therefore, there are no solutions to the given equation.