How do you simplify the expression #cot(arcsin (-7/13))#?

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Answer 1 · 2016-06-24T17:12:39

#-sqrt 120/7# against principal value of arc sin (-7/13), in the 4th quadrant. Of course, the general solution is #+-sqrt120/7#.

Let #a = arc sin (-7/13)#. Then, #sin a = -7/13<0#.

a is in either 3rd quadrant or in the fourth. The principal value is in

4th. Accordingly,

#cos a = sqrt(1-7^2/13^2)=sqrt 120/13#, against principal a.

Also, for the 3rd quadrant a, #cos a=-sqrt120/13#.

The given expression is

#cot a =cos a/sin a=- sqrt120/7#, against principal a and

#sqrt120/7#, for the other a.