How do you calculate # [tan^-1(-1) + cos^-1(-4/5)]#?

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Answer 1 · 2016-05-15T11:42:05

#278.13^o, 351.87^o, 458.13^o and 531.87^o#, choosing the range #[0^o, 360^0]#, for the inverse functions.. .

It is anticlockwise rotation for angles. So, negative angles do not appear.

Let #a = tan^(-1)(-1)#.

Then, #tan a = -1<0#.

So, a is in the 2nd or 4th quadrant.

The solutions in #[0^o, 360^0]# are #a = 135^o and 315^o#.

Let #b = cos^(-1)(-3/4)#.

Then, #cos a = -3/4<0#.

So, a is in the 2nd or 3rd quadrant.

The solutions in #[0^o, 360^0]# are #b = 143.13^o and 216.87^o#

The given expression is

#a + b = 278.13^o, 351.87^o, 458.13^o and 531.87^o# .

If negative angles are used like #[-180^o, 180^0] for the range of

separate solutions, instead, the results would change, accordingly.