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प्रश्न
`4tan^-1 1/5-tan^-1 1/239=pi/4`
उत्तर
LHS = `4tan^-1 1/5-tan^-1 1/239`
`=2tan^-1{(2xx1/5)/(1-(1/5)^2)}-tan^-1 1/239` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=2tan^-1{(2/5)/(24/25)}-tan^-1 1/239`
`=2tan^-1 5/12-tan^-1 1/239`
`=tan^-1{(2xx5/12)/(1-(5/12)^2)}-tan^-1 1/239` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=tan^-1{(5/6)/(119/144)}-tan^-1 1/239`
`=tan^-1 120/119-tan^-1 1/239`
`=tan^-1((120/119-17/239)/(1+120/119xx1/239))` `[becausetan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))]`
`=tan^-1 1=pi/4=`RHS
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