Question

If $cos\theta =\frac{-3}{5}$ and $\pi <\theta <\frac{3\pi }{2}$, then the value of $\frac{cosec\theta +cot\theta }{sec\theta -tan\theta }$ is

• A. $1/6$
• B. $1/7$
• C. $1/5$
• D. $1/2$

Answer 1

Answer: (A)

Given, $cos\theta =\frac{-3}{5}$ and $\pi <\theta <\frac{3\pi }{2}$, therefore
$\theta$ lies in third quadrant. Therefore, in third quadrant
$sin\theta <\theta$, $cos\theta <\theta$ and $tan\theta >\theta$.
We have, $si{n}^2\theta =1-co{s}^2\theta =\frac{16}{25}$
$\Rightarrow sin\theta =\pm \frac{4}{5}$ but $sin\theta <0$.
$\therefore sin\theta =-\frac{4}{5}$;
$sec\theta =\frac{1}{cos\theta }=\frac{-5}{3}$,
$cosec\theta =\frac{1}{sin\theta }=\frac{-5}{4}$
$tan\theta =\frac{sin\theta }{cos\theta }=\frac{4}{3}$ and
$cot\theta =\frac{1}{tan\theta }=\frac{3}{4}$
Now, $\frac{cosec\theta +cot\theta }{sec\theta -tan\theta }$
$=\frac{-\frac{5}{4}+\frac{3}{4}}{-\frac{5}{3}-\frac{4}{3}}$
$=\frac{-2/4}{-9/3}=\frac{1}{6}$