Question

If \[\cos x = - \frac{3}{5}\text{ and }\pi < x < \frac{3\pi}{2}\] find the values of other five trigonometric functions and hence evaluate \[\frac{cosec x + \cot x}{\sec x - \tan x}\]

Answer 1

We have: 
\[\cos x = - \frac{3}{5}\text{ and }\pi < x < \frac{3\pi}{2}\]
Thus, x is in the third quadrant. 
In the third quadrant, tan x and cot x are positive
And, all the other four T - ratios are negative. 
\[ \therefore \sin x = - \sqrt{1 - \cos^2 x} = - \sqrt{1 - \left( \frac{- 3}{5} \right)^2} = \frac{- 4}{5}\]
\[\tan x = \frac{\sin x}{\cos x} = \frac{\frac{- 4}{5}}{- \frac{3}{5}} = \frac{4}{3}\]
\[\cot x = \frac{1}{\tan x} = \frac{1}{\frac{4}{3}} = \frac{3}{4}\]
\[\sec x = \frac{1}{\cos x} = \frac{1}{\frac{- 3}{5}} = \frac{- 5}{3}\]
\[cosec x = \frac{1}{\sin x} = \frac{1}{\frac{- 4}{5}} = \frac{- 5}{4}\]
\[\text{ Now, }\frac{cosec\theta + \cot\theta}{\sec\theta - \tan\theta} = \frac{\frac{- 5}{4} + \frac{3}{4}}{\frac{- 5}{3} - \frac{4}{3}}\]
\[ = \frac{\frac{- 2}{4}}{\frac{- 9}{3}}\]
\[ = \frac{\frac{- 1}{2}}{- 3} = \frac{1}{6}\]