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प्रश्न
`2tan^-1 3/4-tan^-1 17/31=pi/4`
उत्तर
LHS = `2tan^-1 3/4-tan^-1 17/31`
`=tan^-1{(2xx3/4)/(1-(3/4)^2)}-tan^-1 17/31` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=tan^-1{(3/2)/(7/16)}-tan^-1 17/31`
`=tan^-1 24/7-tan^-1 17/31`
`=tan^-1((24/7-17/31)/(1+24/7xx17/31))` `[becausetan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))]`
`=tan^-1((625/217)/(625/217))`
`=tan^-1 1=pi/4=` RHS
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