Find Other Five Trigonometric Functions and Evaluate the Expression

Question:

If

\[ \cos x = -\frac{3}{5} \]

and

\[ \pi < x < \frac{3\pi}{2} \]

find the values of other five trigonometric functions and hence evaluate:

\[ \frac{\cosec x + \cot x}{\sec x – \tan x} \]

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Solution

Given,

\[ \cos x = -\frac{3}{5} \]

Since

\[ \pi < x < \frac{3\pi}{2} \]

\(x\) lies in Quadrant III, where sine and cosine are negative while tangent is positive.

Using:

\[ \cos x = \frac{\text{Base}}{\text{Hypotenuse}} \]

Take,

\[ \text{Base} = -3, \quad \text{Hypotenuse} = 5 \]

Using Pythagoras theorem:

\[ \text{Perpendicular} = \sqrt{5^2 – 3^2} \]

\[ = \sqrt{25 – 9} = \sqrt{16} = 4 \]

In Quadrant III, perpendicular is negative.

Therefore,

\[ \sin x = -\frac{4}{5} \]

\[ \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \]

\[ \sec x = \frac{1}{\cos x} = -\frac{5}{3} \]

\[ \cosec x = \frac{1}{\sin x} = -\frac{5}{4} \]

\[ \cot x = \frac{1}{\tan x} = \frac{3}{4} \]

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Evaluate the Expression

\[ \frac{\cosec x + \cot x}{\sec x – \tan x} \]

Substitute the values:

\[ = \frac{-\frac{5}{4} + \frac{3}{4}}{-\frac{5}{3} – \frac{4}{3}} \]

\[ = \frac{-\frac{2}{4}}{-\frac{9}{3}} \]

\[ = \frac{-\frac{1}{2}}{-3} \]

\[ = \frac{1}{6} \]

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Final Answer

\[ \sin x = -\frac{4}{5} \]

\[ \tan x = \frac{4}{3} \]

\[ \sec x = -\frac{5}{3} \]

\[ \cosec x = -\frac{5}{4} \]

\[ \cot x = \frac{3}{4} \]

and

\[ \boxed{ \frac{\cosec x + \cot x}{\sec x – \tan x} = \frac{1}{6} } \]