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Q.
If $cot\,x=\frac{4}{3}$ and $x$ lies in third quadrant, then find the value of sec $x$.
Trigonometric Functions
Solution:
Since, $sec^{2}x=1+tan^{2}\,x$
$=1+\left(\frac{3}{4}\right)^{2}=\frac{25}{16}$
On taking square root, we get $sec\,x=\pm\frac{5}{4}$
But $x$ lies in third quadrant, so $sec \,x=-\frac{5}{4}$