Evaluate arctan5/12-arccos3/5

We'll put a = arctan 5/12 and b =
arccos3/5.

So, we'll have to compute the difference between
angles:

a - b

We'll apply the
sine function to the difference above:

sin (a-b) = sin
a*cos b - sin b*cos a

Since a = arctan 5/12, then tan a =
5/12.

Also, b = arccos3/5, then cos b =
3/5.

We'll apply the Pythagorean Theorem and we'll
get:

(sin a)^2 + (cos a)^2 =
1

We'll divide by (cos a)^2 both sides and since tan a =
sin a/cos a

(tan a)^2 + 1 = 1/(cos
a)^2

(cos a)^2 = 1/[(tan a)^2 +
1]

(cos a)^2 = 1/(25/144 +
1)

(cos a)^2 =
144/(25+144)

(cos a)^2 =
144/169

cos a = +/-
12/13

Since tan a = sin a/cos a =>
sin a = tan a*cos a

sin a =
(5/12)*(12/13)

sin a =
+/-5/13

From the funcdamental formula of
trigonometry, we'll calculate sin b.

cos b =
3/5

(sin b)^2 + (cos b)^2 =
1

(sin b)^2 = 1 - (cos
b)^2

sin b = sqrt(1 -
9/25)

sin b =
sqrt[(25-9)/25]

sin b
=+/-4/5

Now, we can compute sin
(a-b):

sin (a-b)=(5/13)*(3/5) -
(4/5)(12/13)

sin (a-b) = (15 -
48)/5*13

sin (a-b) =
-33/65

arctan 5/12 - arccos3/5
= -33/65