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प्रश्न
If `tan^-1((x-1)/(x-2))+cot^-1((x+2)/(x+1))=pi/4; `
उत्तर
`"Given ",tan^-1((x-1)/(x-2))+cot^-1((x+2)/(x+1))=pi/4`
`tan^-1((x-1)/(x-2))+tan^-1((x+1)/(x+2))=pi/4;`
`tan^-1((x-1)/(x-2))=pi/4-tan^-1((x+1)/(x+2))`
`=tan^-1(1)-tan^-1((x+1)/(x+2))`
`=tan^-1[(1-(x+1)/(x+2))/(1+(x+1)/(x+2))] ........[because tan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))]`
`=tan^-1[(x+2-x-1)/(x+2+x+1)]`
`tan^-1((x-1)/(x-2))=tan^-1(1/(2x+3))`
`(x-1)/(x-2)=1/(2x+3)`
`(x-1)(2x+3)=x-2`
`2x^2-1=0`
`2x^2=1`
`x^2=1/2`
`x=+-1/sqrt2`
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