Question
Prove `sin^(-1) 8/17 + sin^(-1) 3/5 = tan^(-1) 77/36`
Solution
Let sin^(-1) 8/17 = x. Then `sinx = 8/17 => cos x = sqrt(1 - (8/17)^2) = sqrt(225/289) = 15/17`
`:. tanx = 8/15 => x = tan^(-1) 8/15`
`:. sin^(-1) 8/17 = tan^(-1) 8/15` ....(1)
Now let `sin^(-1) 3/5 = y. " Then " sin y = 3/5 => cos y = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5`
`:. tan y = 3/4 => y = tan^(-1) 3/4`
`:. sin^(-1) 3/5 = tan^(-1) 3/4` .... 2
Now, we have:
L.H.S = `sin^(-1) 8/17 + sin^(-1) 3/5`
`= tan^(-1) 8/15 + tan^(-1) 3/4 ` [Using 1 and 2]
`= tan^(-1) (8/15 + 3/4)/(1 - 8/15 xx 3/4)`
`= tan^(-1) ((32+45)/(60 - 24)) ` `" "[tan^(-1) x + tan^(-1) y = tan^(-1) (x+y)/(1-xy)] `
`= tan^(-1) 77/36` = R.H.S
Is there an error in this question or solution?
Solution Prove `Sin^(-1) 8/17 + Sin^(-1) 3/5 = Tan^(-1) 77/36` Concept: Properties of Inverse Trigonometric Functions.