Answer 1

Let a = `"cos"^-1 (4/5)` and b = `"cos"^-1 (12/13)`

Let a = `"cos"^-1 (4/5)`

cos a = `4/5`

We know that

sin²a = 1 - cos²a 

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

`= sqrt (1 - (4/5)^2) = sqrt (1 - 16/25)`

`= sqrt ((25-16)/25) = sqrt (9/25) = 3/5`

Let b = `"cos"^-1 (12/13)`

cos b = `12/13`

W know that

sin²b = 1 - cos²b 

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

`= sqrt (1 - (12/13)^2) = sqrt (1 - 144/169)`

`= sqrt ((169-144)/169) = sqrt (25/169) = 5/13`

We know that 

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

Putting values 

cos a = `4/5` , sin a = `3/5`

& cos b = `12/13` , sin b = `5/13`

 

cos (a+b) = `4/5 xx 12/13 xx 3/5 xx 5/13`

`= 48/65 - 3/13`

`= (48 - 15)/65`

`= 33/65`

∴ cos (a+b) = `33/65`

a + b = cos^-1 `(33/65)`

`"cos"^-1 4/5 + "cos"^-1 (12/15) = "cos"^-1 (33/65)`

Hence LH.S = R.H.S

Hence proved.