Find the value of ![image.png](https://res.cloudinary.com/aptidude/image/upload/v1746630255/aptidude_questions/w0u4fdeyyzzn83grpblk.png)

Question

Accepted Answer

Let’s evaluate the given expression step by step for $$A = 60^\circ$$:

**Step 1: Find values of trigonometric functions at $$60^\circ$$:**  
$$
\csc 60^\circ = \frac{2}{\sqrt{3}}, \quad \sec 60^\circ = 2
$$

**Step 2: Substitute values in the expression:**  
First part:
$$
\frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1}
$$
Let $$x = \csc A$$:
$$
= \frac{x}{x-1} + \frac{x}{x+1}
= \frac{x(x+1) + x(x-1)}{(x-1)(x+1)}
= \frac{x^2 + x + x^2 - x}{x^2 - 1}
= \frac{2x^2}{x^2 - 1}
$$

For $$x = \frac{2}{\sqrt{3}}$$:
$$
x^2 = \frac{4}{3}
$$
$$
\frac{2x^2}{x^2 - 1} = \frac{2 	imes \frac{4}{3}}{\frac{4}{3} - 1} = \frac{\frac{8}{3}}{\frac{1}{3}} = 8
$$

Second part:
$$
\frac{\sec A}{\sec A - 1} + \frac{\sec A}{\sec A + 1}
$$
Let $$y = \sec A$$:
$$
= \frac{y}{y-1} + \frac{y}{y+1}
= \frac{y(y+1) + y(y-1)}{(y-1)(y+1)}
= \frac{y^2 + y + y^2 - y}{y^2 - 1}
= \frac{2y^2}{y^2 - 1}
$$

For $$y = 2$$:
$$
y^2 = 4
$$
$$
\frac{2y^2}{y^2 - 1} = \frac{2 	imes 4}{4 - 1} = \frac{8}{3}
$$

**Step 3: Combine and apply exponent:**  
$$
\left[8 	imes \left(\frac{8}{3}ight)^{-1}ight] = 8 	imes \frac{3}{8} = 3
$$

**Final Answer:**  
The value of the given expression when $$A = 60^\circ$$ is **3**.

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