Question : If $\sin A=-\frac{3}{5}, A$ lies in III quadrant, the value of $\sec A$ is:
Option 1: $-\frac{4}{5}$
Option 2: $-\frac{5}{4}$
Option 3: $-\frac{3}{4}$
Option 4: $\frac{3}{4}$
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Correct Answer: $-\frac{5}{4}$
Solution : $\sin A=-\frac{3}{5}$ So, $\cos A = \sqrt{1-\sin^2 A}$ ⇒ $\cos A= \sqrt{1-(\frac{3}{5})^2}$ ⇒ $\cos A = \sqrt\frac{16}{25}$ $\therefore\cos A= \pm \frac{4}{5}$ Since it is 3rd quadrant, $\cos A = -\frac{4}{5}$ $\therefore\sec A = -\frac{5}{4}$ Hence, the correct answer is $-\frac{5}{4}$.
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