Question

If x lies in third quadrant and 5 sin x + 3 = 0, find the value of

$\frac{2tanx-5sinx+cotx}{2sin\frac{x}{2}cos\frac{x}{2}}$

Answer 1

We have, 5 sin x + 3 = 0 $\Rightarrow sinx=-\frac{3}{5}$

$\because si{n}^{2}x+co{s}^{2}x=1$

$\therefore cosx=-\frac{4}{5}$

[$\because$ x is in IIIrd quadrant, so we take negative sign for cos x]

$\therefore tanx=\frac{sinx}{cosx}=\frac{3}{4}$

[$\because$ x is in IIIrd quadrant, so we take negative sign for tan x]

Now, $\frac{2tanx-5sinx+cotx}{2sin\frac{x}{2}cos\frac{x}{2}}$

$=\frac{2\times \frac{3}{4}-5\times (-\frac{3}{5})+\frac{4}{3}}{sinx}$

$[\because sin2\theta =2sin\theta cos\theta ]$

$=\frac{\frac{3}{2}+3+\frac{4}{3}}{-\frac{3}{5}}=\frac{\frac{9-18+8}{6}}{-\frac{3}{5}}$

$=-\frac{35\times 5}{6\times 3}=-\frac{175}{18}$