How do you solve and find the value of #cot(sin^-1(5/6))#?

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Answer 1 · 2017-01-19T14:21:33

#cottheta=+sqrt11/5#.

Let #sin^-1(5/6)=theta rArr sintheta=5/6, theta in [-pi/2,pi/2]#

But as #sintheta >0, theta in (0,pi/2)#

We require the value=#cottheta#

Knowing that #csc^2theta=1+cot^2theta, we have,

#cot^2theta=csc^2theta-1=1/sin^2theta-1=36/25-1=11/25#

#rArr cottheta=+-sqrt11/5#

As, #theta in (0,pi/2), cottheta=+sqrt11/5#.