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Find the values of other five trigonometric functions tan x = - 5/12 , x lies in the second quadrant

Q (5) Find the values of the other five trigonometric functions

\small \tan x = -\frac{5}{12}  , x lies in second quadrant.

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Solution 
 \tan x = -\frac {5}{12}
\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}
1 + \tan^{2}x = \sec^{2}x\\ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x\\ 1 + \frac {25}{144} = \sec^{2}x\\ \\ \frac {169}{144} = \sec^{2}x\\ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}
x lies in second quadrant. Therefore the value of sec x is negative
\sec x = - \frac {13}{12}
\cos x = \frac{1}{\sec x}= \frac{1}{-\frac{13}{12}} = -\frac {12}{13}
\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \left ( -\frac{12}{13} \right )^{2}\\ \sin^{2}x = 1 - \frac{144}{169}\\ \sin^{2}x = \frac {25}{169}\\ \sin x = \sqrt {\frac{25}{169}} = \pm \frac{5}{13}
x lies in the second quadrant. Therefore the value of sin x is positive
\sin x = \frac {5}{13}
\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}

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