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.

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