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Q. If $\cos x=-\frac{3}{5}$ and $x$ lies in third quadrant, then which among the following is/are correct?
I. $\sin x=-\frac{4}{5}$
II. $\operatorname{cosec} x=\frac{-5}{4}$
III. $ \sec x=\frac{-5}{3} $
IV.$ \cot x=\frac{4}{3}$
V. $\tan x=\frac{3}{4}$

Trigonometric Functions

Solution:

Since, $\cos x=-\frac{3}{5}$, we have $\sec x=\frac{-5}{3}$
Now, $\sin ^2 x+\cos ^2 x=1$, i.e., $\sin ^2 x=1-\cos ^2 x$
$\Rightarrow \sin ^2 x=1-\frac{9}{25}=\frac{16}{25}$
Hence, $ \sin x=\pm \frac{4}{5}$
Since, $x$ lies in third quadrant, $\sin x$ is negative.
Therefore,
$\sin x=-\frac{4}{5}$
which also gives
$\operatorname{cosec} x=-\frac{5}{4}$
Further, we have
$\tan x =\frac{\sin x}{\cos x}=\frac{4}{3} $
and $ \cot x=\frac{\cos x}{\sin x}=\frac{3}{4}$