# Given secphi=3/2, cscphi=-(3sqrt5)/5 to find the remaining trigonometric function?

Feb 25, 2018

We know, $\cos \phi = \frac{1}{\sec \phi} = \frac{2}{3}$

Again, ${\sin}^{2} \phi + {\cos}^{2} \phi = 1$

so, ${\sin}^{2} \phi = 1 - {\left(\frac{2}{3}\right)}^{2} = \frac{5}{9}$

So, $\sin \phi {=}_{-}^{+} \frac{\sqrt{5}}{3}$

So, $\csc \phi = \frac{1}{\sin \phi} {=}_{-}^{+} \frac{3}{\sqrt{5}}$

But, negative value of $\csc \phi$ is considered only,so, $\sin \phi = - \frac{\sqrt{5}}{3}$

So, $\tan \phi = \sin \frac{\phi}{\cos} \phi = - \frac{\sqrt{5}}{2}$

Hence, $\cot \phi = - \frac{2}{\sqrt{5}}$