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

Explanation

Let ` sin^(-1) ((5)/(13)) = x `
` therefore sin x = (5)/(13)` …(i)
But ` sin^(2) x + cos^(2) x = 1`
` therefore (25)/(169) + cos^(2) x = 1`
`rArr cos^(2) x = 1 - (25)/(169) = (144)/(169) `
` rArr therefore cos x = (12)/(13) ` ...(ii)
and let ` cos^(-1)((3)/(5)) = y `
` therefore cos y = (3)/(5) ` ...(iii)
But ` sin^(2) y + cos^(2) y = 1`
` therefore sin^(2) y + cos^(2) y = 1`
` therefore sin^(2) y + (9)/(25) = 1`
` rArr therefore sin^(2) y = 1 - (9)/(25) = (16)/(25)`
`rArr therefore sin y = (4)/(5) ` ...(iv)
Now , ` sin (x + y ) = sin x . cos y + cos x . sin y `
` = (5)/(13) , (3)/(5) + (12)/(13) . (4)/(5)`
` = (3)/(13) + (12xx 4)/(13xx5)`
` = (1)/(13) (3 + (48)/(5))`
` therefore sin (x + y) = (63)/(65)` ...(v)
and ` cos (x +y) = cos x . cos y - sin x . sin y `
` = (12)/(12) , (3)/(5) - (5)/(13) , (4)/(5)`
` = (36)/(65) - (20)/(65) = (36 - 20)/(65) `
` cos (x + y ) = 16/65 `... (vi)
Now , ` tan (x + y) = (sin (x + y))/(cos x + y)`
`=((63)/(65))/((16)/(65)) `
` therefore tan (x + y) = (63)/(16) `
` therefore x + y = tan^(-1) ((63)/(16))`
` therefore ` From equation (i) and (iii) we get ,
`sin^(-1) ((5)/(13)) + cos^(-1) ((3)/(5)) = tan^(-1) ((63)/(16))` .