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  4. If cosθ=(-3/5) and π < θ <(3π/2), then the value of (cosecθ+cot θ/sec θ-tan θ) is

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Q. 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

Trigonometric Functions

Solution:

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, $sin^{2}\theta=1-cos^{2}\,\theta=\frac{16}{25}$
$\Rightarrow sin\theta=\pm \frac{4}{5}$ 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, $\frac{cosec\theta+cot\,\theta}{sec\,\theta-tan\,\theta}$
$=\frac{-\frac{5}{4}+\frac{3}{4}}{-\frac{5}{3}-\frac{4}{3}}$
$=\frac{-2/4}{-9/3}=\frac{1}{6}$