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प्रश्न
`tan^-1 1/7+2tan^-1 1/3=pi/4`
उत्तर
LHS = `tan^-1 1/7+2tan^-1 1/3`
`=tan^-1 1/7+tan^-1{(2xx1/3)/(1-(1/3)^2)}` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=tan^-1 1/7+tan^-1{(2/3)/(8/9)}`
`=tan^-1 1/7+tan^-1 3/4`
`=tan^-1((1/7+3/4)/(1-1/7xx3/4))` `[becausetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`
`=tan^-1((25/28)/(25/28))`
`=tan^-1 1=pi/4=`RHS
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