If cot x = -frac{5}{12} and x lies in the sec | vedclass

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Question

(A) Given $\cot x = -\frac{5}{12}$. Since $x$ is in the second quadrant,$	an x = \frac{1}{\cot x} = -\frac{12}{5}$.Using the identity $\sec^2 x = 1 +

Vedclass · Accepted Answer

$\sin x = \frac{12}{13}, \cos x = -\frac{5}{13}, 	an x = -\frac{12}{5}, \csc x = \frac{13}{12}, \sec x = -\frac{13}{5}$

Vedclass · Answer

(A) Given $\cot x = -\frac{5}{12}$. Since $x$ is in the second quadrant,$	an x = \frac{1}{\cot x} = -\frac{12}{5}$.Using the identity $\sec^2 x = 1 + 	an^2 x = 1 + (-\frac{12}{5})^2 = 1 + \frac{144}{25} = \frac{169}{25}$.Since $x$ is in the second quadrant,$\sec x$ must be negative,so $\sec x = -\frac{13}{5}$.Consequently,$\cos x = \frac{1}{\sec x} = -\frac{5}{13}$.Now,$\sin x = 	an x \cdot \cos x = (-\frac{12}{5}) \cdot (-\frac{5}{13}) = \frac{12}{13}$.Finally,$\csc x = \frac{1}{\sin x} = \frac{13}{12}$.

If $\cot x = -\frac{5}{12}$ and $x$ lies in the second quadrant,find the values of the other five trigonometric functions.

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