Answer 1

Let `sin^-1  3/5 = "x"` 
Then, `sin"x" = 3/5 ⇒ cos"x" = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5`

`∴ tan"x" = 3/4 ⇒ "x" = tan^-1  3/4`

`∴ sin ^-1  3/5 = tan^-1  3/4` ......(1)

Now,let `cos^-1  12/13 = "y".`
Then, `cos"y" = 12/13 ⇒ sin"y" = 5/13`

`∴ tan"y" = 5/12 ⇒ "y" = tan^-1  5/12`

`∴cos^-1  12/13 = tan^-1  5/12`  ........(2)

Let `sin^-1  56/65 = "z".`
Then, `sin"z" = 56/65 ⇒ cos"z" = 33/65`

`∴ tan"z" = 56/33 ⇒ "z" = tan^-1  56/33`

`∴sin^-1  56/65 = tan^-1  56/33`   .........(3)

Now, we have:

L.H.S. = `cos^-1  12/13 + sin^-1  3/5`

`= tan^-1  5/12 + tan^-1  3/4`  .........[Ueing (1) and (2)]

`=tan^-1  (5/12+3/4)/(1-5/12 . 3/4)`  ........`[tan^-1"x"+ tan^-1"y"=tan^-1  "x+y"/(1-"xy")]`

` = tan^-1  (20+36)/(48-15)`

`= tan^-1  56/33`

`= sin^-1  56/65 = "R.H.S"`  .........[Using (3)]