Prove: `tan^(-1)(1/4)+tan^(-1)(2/9)=sin^(-1)(1/sqrt(5))`

Answer

N/a

Explanation

We have, `tan^(-1)'1/4+tan^(-1)'2/9=sin^(-1)'1/(sqrt(5))`.
Let `tan^(-1)'1/4 =x`
`rArr tanx = 1/4`
`rArr tan^(2)x=1/16`
`rArr sec^(2)x-1=1/16`
`rArr sec^(2)x=1+1/16=17/16`
`rArr 1/(cos^(2)x)=17/16`
`rArr cos^(2)x = 16/17`
`rArr cosx = 4/(sqrt(17))`
`rArr sin^(2)x=1-cos^(2)x=1-16/17=1/17`
`rArr sinx = 1/(sqrt(17))`
Again, let `tan^(-1)'2/9 = y`
`rArr tany = 2/9 rArr tan^(2)y = 4/81`
`rArr sec^(2)y - 1 = 4/81`
`rArr sec^(2)y = 4/81 + 1 = 85/81`
`rArr cos^(2)y = 81/85 rArr cosy = 8/(sqrt(85))`
`rArr sin^(2)y = 1- cos^(2)y = 1 - (81)/(85) = 4/85`
`rArr siny = 8/(sqrt(85))"........"(iii)`
We know that, `sin(x+y)=sinx.cosy+cosx.siny`
`= 1/(sqrt(17)).(9)/(sqrt(85)) + 4/(sqrt(17)).(2)/(sqrt(85))`
`= (17)/(sqrt(17).sqrt(85)) = (sqrt(17))/(sqrt(17).sqrt(5)) = 1/(sqrt(5))`
`rArr (x+y) = sin^(-1)'(1)/(sqrt(5))`
`rArr tan^(-1)'(1)/(4) + tan^(-1)'(2)/(9)sin^(-1)'(1)/(sqrt(5))`