If sin x = -frac{3}{5},where pi

Question

(A) Given $\sin x = -\frac{3}{5}$ and $\pi < x < \frac{3\pi}{2}$ (which is the third quadrant).In the third quadrant,$	an x$ is positive and $\cos x$

Vedclass · Accepted Answer

$109$

Vedclass · Answer

(A) Given $\sin x = -\frac{3}{5}$ and $\pi In the third quadrant,$	an x$ is positive and $\cos x$ is negative.Using $\cos^2 x = 1 - \sin^2 x$,we get $\cos^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.Since $x$ is in the third quadrant,$\cos x = -\frac{4}{5}$.Now,$	an x = \frac{\sin x}{\cos x} = \frac{-3/5}{-4/5} = \frac{3}{4}$.Therefore,$	an^2 x = (\frac{3}{4})^2 = \frac{9}{16}$.Substituting these values into the expression $80(	an^2 x - \cos x)$:$80(\frac{9}{16} - (-\frac{4}{5})) = 80(\frac{9}{16} + \frac{4}{5})$.$= 80(\frac{45 + 64}{80}) = 45 + 64 = 109$.

If $\sin x = -\frac{3}{5}$,where $\pi < x < \frac{3\pi}{2}$,then $80(\tan^2 x - \cos x)$ is equal to:

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