Question

Prove the following result-

`tan^-1  63/16 = sin^-1  5/13 + cos^-1  3/5`

Answer 1

Let a = `sin^-1  5/13` b = `cos^-1  3/5`

Let a = `sin^-1  5/13`

We know that

cos a = `sqrt(1 - sin^2 "a")`

`= sqrt(1 - (5/13)^2)`

`= sqrt(144/169)`

`= 12/13`

Let b = `cos^-1  3/5`

cos b = `3/5`

We know that

sin b = `sqrt(1 - cos^2 "b")`

`= sqrt(1 - (3/5)^2)`

`= sqrt(16/25)`

`= 4/5`

Now, 

tan a = `(sin a)/(cos a)`

`= (5/13)/(12/13)`

`= 5/13 xx 13/12`

`= 5/12`

tan b = `(sin b)/(cos b)`

`= (4/5)/(3/5)`

`= 4/5 xx 5/3`

`= 4/3`

Now we know that

tan (a + b) = `(tan a + tan b)/(1 - tan a  tan b)`

Putting tan a = `5/12` and tan b = `4/3`

tan (a + b) = `(5/12 + 4/3)/(1 - 5/12 xx 4/3)`

tan (a + b) = `((5 xx 3 + 4 xx 12)/36)/(1 - 20/36)`

= `((15 + 48)/36)/((36 - 20)/36)`

= `(63/36)/(16/36)`

= `63/36 xx 36/16`

= `63/16`

Thus, tan (a + b) = `63/16`

a + b = `tan^-1 (63/16)`

Putting values of a and b

`sin^-1  5/13 + cos^-1  3/5 = tan^-1  (63/16)`

Hence L.H.S = R.H.S

Hence Proved.