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: $\frac{3}{4}$
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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