Question

If $cos\theta =\frac{3}{5}$,and $\pi <\theta <\frac{3\pi }{2}$, find the values of other five trigonometric functions and hence evaluate $\frac{cosec\theta +cot\theta }{sec\theta -tan\theta }$.

Answer 1

We have,

$cos\theta =-\frac{3}{5}$, and $\pi <\theta <\frac{3\pi }{5}$

$\Rightarrow \theta$ lies in the 3rd quadrant

We know that,

$\Rightarrow sin\theta =\pm \sqrt{1-co{s}^2\theta }$

In the 3rd quadrant sin $\theta$ is negative and $tan\theta$ is positive.

$\therefore sin\theta =-\sqrt{1-co{s}^2\theta }$

$=-\sqrt{1-(-\frac{3}{5}{)}^2}[\because cos\theta =-\frac{3}{5}]$

$=-\sqrt{1-\frac{9}{25}}$

$=-\sqrt{\frac{16}{25}}$

$\Rightarrow sin\theta =-\frac{4}{5}$

and, $tan\theta =\frac{sin\theta }{cos\theta }=\frac{\frac{-4}{5}}{\frac{-3}{5}}=\frac{4}{5}$

Now, $cosec\theta =\frac{1}{sin\theta }=\frac{1}{\frac{-4}{5}}=\frac{-5}{4}$

$sec\theta =\frac{1}{cos\theta }=\frac{1}{\frac{-3}{5}}=\frac{-5}{3}$

and, $cot\theta =\frac{1}{tan\theta }=\frac{1}{\frac{4}{3}}=\frac{3}{4}$

$\therefore \frac{cosec\theta +cot\theta }{sec\theta -tan\theta }=\frac{\frac{-5}{4}+\frac{3}{4}}{\frac{-5}{3}-\frac{4}{3}}$

$=\frac{\frac{-5+3}{4}}{\frac{-5-4}{3}}$

$=\frac{\frac{-2}{4}}{\frac{9}{-3}}=\frac{2}{4}\times \frac{3}{9}=\frac{1}{6}$

$\therefore \frac{cosec\theta +cot\theta }{sec\theta -tan\theta }=\frac{1}{6}$