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प्रश्न
Solve the following equation for x:
`tan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
उत्तर
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
`thereforetan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
`=>tan^-1 1/4+tan^-1 1/5+tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
`=>tan^-1((1/4+1/5)/(1-1/4xx1/5))+tan^-1((1/5+1/6)/(1-1/5xx1/6))+tan^-1 1/x=pi/4`
`=>tan^-1((9/20)/(19/20))+tan^-1((11/30)/(29/30))+tan^-1 1/x=pi/4`
`=>tan^-1(9/19)+tan^-1(11/29)+tan^-1 1/x=pi/4`
`=>tan^-1((9/19+11/29)/(1-11/29xx1/x))+tan^-1 1/x=pi/4`
`=>tan^-1 (235/226)+tan^-1 1/x=pi/4`
`=>tan^-1((235/226+1/x)/(1-235/226xx1/x))=pi/4`
`=>(235x+226)/(226x-235)=tan pi/4`
`=>(235x+226)/(226x-235)=1`
`=>235x+226=226x-235`
`=>9x=-461`
`=>x=-461/9`
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