To evaluate arctan5/12 -arccos3/5

If arc cos 3/5 = y,

3/5 = cosy.

So tany = siny/cosy

tany = sqrt(1-cos^2y)/cosy

tany = sqrt(1-(3/5)^2)/cosy.

tany ={ sqrt(25-9)/5}/(3/5) = 4/3.

Therefore y = arctan (4/3).

Therefore arc tan (5/12) - arccos (3/5) = arctan (5/12)- arctan (4/3).

Therefore , tan {arctan 5/12 - arctan 4/3}

= {5/12 - 4/3}/(1+{5/12)(4/3)} = 3(5-16)/{36+20)

tan {arctan 5/12 - arctan 4/3} = -33/56.

Therefore arctan 5/12 - arctan 4/3}= -arctan (33/56)

arctan 5/12 - arctan 4/3 = -30.51 deg

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**