If \[ \frac{\sec \alpha + \tan \alpha}{\sec \alpha - \tan \alpha} = \frac{7}{4}, \] then find the value of \( \csc \alpha \) .

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SSC CGL - 2024

If \[ \frac{\sec \alpha + \tan \alpha}{\sec \alpha - \tan \alpha} = \frac{7}{4}, \] then find the value of \( \csc \alpha \).

This question was previously asked in

SSC CGL 2024 (Tier-I) Previous Year Paper (17-Sep-2024) (Shift 3)

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11/3

The correct answer is

11/3

Given:

\[\frac{\sec \alpha + \tan \alpha}{\sec \alpha - \tan \alpha} = \frac{7}{4}\]

Let:

\[x = \sec \alpha, \quad y = \tan \alpha\]

Then the equation becomes:

\[\frac{x + y}{x - y} = \frac{7}{4}\]

Cross-multiplying:

\[4(x + y) = 7(x - y)\]

Expanding both sides:

\[4x + 4y = 7x - 7y\]

Bringing all terms to one side:

\[4x + 4y - 7x + 7y = 0 \Rightarrow -3x + 11y = 0 \Rightarrow 11y = 3x \Rightarrow \frac{y}{x} = \frac{3}{11}\]

Recall the identity:

\[\frac{\tan \alpha}{\sec \alpha} = \sin \alpha \Rightarrow \sin \alpha = \frac{3}{11}\]

Therefore:

\[\csc \alpha = \frac{1}{\sin \alpha} = \frac{11}{3}\]

Final Answer:

\[\boxed{\csc \alpha = \frac{11}{3}}\]

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