Q (4) Find the values of other five trigonometric functions

\small \sec x = \frac{13}{5}  , x lies in fourth quadrant.

Answers (1)
S safeer

Solution
\sec x = \frac {13}{5}
\cos x = \frac {1}{\sec x} = \frac{1}{\frac {13}{5}} = \frac {5}{13}
\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \frac {5}{13}\\ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}\\ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}
 lies in fourth quadrant. Therefore sin x is negative
\sin x =- \frac {12}{13}
\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12} 
\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}
\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}

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