Question

If $cotx=\frac{4}{3}$ and $x$ lies in third quadrant, then find the value of sec $x$.

• A. $1/4$
• B. $7/4$
• C. $2/4$
• D. $-5/4$

Answer 1

Answer: (D)

Since, $se{c}^{2}x=1+ta{n}^{2}x$
$=1+{\left(\frac{3}{4}\right)}^2=\frac{25}{16}$
On taking square root, we get $secx=\pm \frac{5}{4}$
But $x$ lies in third quadrant, so $secx=-\frac{5}{4}$