Exams › SSC CGL (Prelims) › General

If \(\cos A = \frac{m}{n}\), then what is the value of \(1 + \cot^2 A\)?

1. \(\frac{n^2}{n^2 - m^2}\)
2. \(\frac{2n^2}{n^2 - m^2}\)
3. \(\frac{n^2}{2(n^2 - m^2)}\)
4. \(\frac{5n^2}{n^2 - m^2}\)

Correct answer: \(\frac{n^2}{n^2 - m^2}\)

Solution

Using the identity \(1+\cot^2 A=\csc^2 A\). Given \(\cos A=\frac{m}{n}\), we get \(\sin^2 A=1-\cos^2 A=1-\frac{m^2}{n^2}=\frac{n^2-m^2}{n^2}\). Hence \(\csc^2 A=\frac{1}{\sin^2 A}=\frac{n^2}{n^2-m^2}\).

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