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प्रश्न
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
उत्तर
LHS = `2tan^-1 1/5+tan^-1 1/8`
`=tan^-1{(2xx1/5)/(1-(1/5)^2)}+tan^-1 1/8` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=tan^-1{(2/5)/(24/25)}+tan^-1 1/8`
`=tan^-1 5/12+tan^-1 1/8`
`=tan^-1((5/12+1/8)/(1-5/12xx1/8))` `[becausetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`
`=tan^-1((13/24)/(91/96))`
`=tan^-1 4/7`= RHS
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