1.

Prove that `2tan^(-1)(1/2)+tan^(-1)(1/7)=sin^(-1)((31)/(25sqrt(2)))`

Answer» `L.H.S. = 2tan^-1(1/2)+tan^-1(1/7)`
`=tan^-1((2*1/2)/(1-(1/2)^2))+tan^-1(1/7)`
`=tan^-1((1)/(3/4))+tan^-1(1/7)`
`=tan^-1(4/3)+tan^-1(1/7)`
`=tan^-1((4/3+1/7)/(1-(4/3)(1/7)))`
`=tan^-1((31/21)/(17/21))`
`=tan^-1(31/17)`
Let `tan^-1(31/17) = theta`
Then, `tantheta = 31/17`
`cot theta = 17/31`
`cosectheta = sqrt(1+(17/31)^2) = sqrt1250/31 = (25sqrt2)/31`
`:. sintheta = 31/(25sqrt2)`
`=>theta = sin^-1(31/(25sqrt2))`
`:. tan^-1(31/17) = sin^-1(31/(25sqrt2)) = R.H.S.`


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