Question

If $A$ lies in the second quadrant and $3tanA+4=0$, then the value of $2cotA-5cosA+siaA$ is equal to

• A. $\frac{-53}{10}$
• B. $\frac{23}{10}$
• C. $\frac{37}{10}$
• D. $\frac{7}{10}$

Answer 1

Answer: (B)

We have, $3tanA+4=0$
or $tanA=-\frac{4}{3}$
$\therefore AC=\pm \sqrt{{\left(-4\right)}^2+{\left(3\right)}^2}$
$=\pm \sqrt{25}=\pm 5$
$\therefore cotA=\frac{-3}{4}$,
$cosA=\frac{-3}{5}$,
$sinA=\frac{4}{5}$
($\because$ Only sin and $cosec$ are $+ve$ in $II$ quadrant)
$\therefore 2cotA-5cosA+sinA$
$=2\times \left(\frac{-3}{4}\right)-5\left(\frac{-3}{5}\right)+\frac{4}{5}$
$=\frac{-3}{2}+3+\frac{4}{5}$
$=\frac{-15+30+8}{10}=\frac{23}{10}$