Prove that `sin^(- 1)(8/17)+sin^(- 1)(3/5)=sin^(- 1)(77/85)`

Answer

N/a

Explanation

We have, `sin^(-1)'(8)/(17)+sin^(-1)'3/5=sin^(-1)'77/85`
`:. LHS = sin^(-1)'8/17+sin^(-1)'3/5`
`= tan ^(-1)'8/15+tan^(-1)'3/4`
Let `sin^(-1)'8/17 = theta_(1) rArr sintheta_(1)= 8/17`

`rArr tantheta_(1) = 8/15 rArr theta_(1) = tan^(-1)'8/15`
and `sin^(-1)'3/5 = theta_(2) rArr sin theta_(2) = 3/5`
`rArr tan theta_(2) = 3/4 rArr theta_(2) = tan^(-1)'3/4`
`tan^(-1)[(8/15+3/4)/(1-8/15xx3/4)] , [:' tan^(-1)x+tan^(-1)y = tan^(-1)'((x+y)/(1-xy))]`
`=tan^(-1)[((32+45)/(60))/((60-24)/60)] = tan^(-1)(77/36)`
Let `theta_(3) = tan^(-1)'77/36 rArr tantheta_(3)=77/36`
`rArr sintheta_(3)=(77)/(sqrt(5929+1296))=77/85`
`:. theta_(3)= sin^(-1)'77/85`
`=sin^(-1)'77/85 = RHS`

Alternate Method
To prove, `sin^(-1)'8/17 + sin^(-1)'3/5 = sin^(-1)'77/85`
Let `sin^(-1)'8/17 = x`
`rAss sinx = 8/17`
`rArr cosx = sqrt(1-sin^(2)theta) = sqrt(1-(8/17)^(2))`
`= sqrt((289-64)/(289)) = sqrt((225)/(289)) = 15/17`

Let `sin^(-1)'3/5 = y`
`rArr sin y = 3/5 rArr sin^(2)y = 9/25`
`rArr cos^(2)y = (4/5)^(2) rArr cos y = 4/5`

Now, `sin(x+y)=sinx.cosy+cosx.siny`
`= 7/17.4/5+15/17.3/5`
` = 32/85+45/85=77/85`
`rArr (x+y) = sin^(-1)(77/85)`
`rArr sin^(-1)'8/17 + sin^(-1)'3/5 = sin^(-1)'77/85`