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Q.
If $A$ lies in the second quadrant and $3tanA + 4 = 0$, then the value of $2cotA - 5cosA + siaA$ is equal to
Trigonometric Functions
Solution:
We have, $3tanA + 4 = 0$
or $tanA=-\frac{4}{3}$
$\therefore AC= \pm\sqrt{\left(-4\right)^{2}+\left(3\right)^{2}}$
$=\pm\sqrt{25} =\pm5$
$\therefore cotA=\frac{-3}{4}$,
$cosA=\frac{-3}{5}$,
$sinA=\frac{4}{5}$
($\because$ Only sin and $cosec$ are $+ve$ in $II$ quadrant)
$\therefore 2cotA-5cosA+sinA$
$=2\times\left(\frac{-3}{4}\right)-5\left(\frac{-3}{5}\right)+\frac{4}{5}$
$=\frac{-3}{2}+3+\frac{4}{5}$
$=\frac{-15+30+8}{10}=\frac{23}{10}$
