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

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Answer 1 · 2018-02-25T16:45:25

We know, # cos phi = 1/(sec phi)=2/3#

Again, #sin^2 phi + cos ^2 phi =1#

so, #sin^2 phi = 1 -(2/3)^2 = 5/9#

So, # sin phi = _-^+ sqrt(5)/3#

So, #csc phi = 1/(sin phi) = _-^+3/sqrt(5)#

But, negative value of #csc phi# is considered only,so, #sin phi = -sqrt(5)/3#

So, #tan phi = sin phi / cos phi=-sqrt(5)/2#

Hence, #cot phi = -2/sqrt(5)#