Prove that :
`cos^-1 (12/13) + sin^-1(3/5) = sin^-1(56/65)`
Solution
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)]