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प्रश्न
Solve the following equation for x:
`tan^-1 x/2+tan^-1 x/3=pi/4, 0<x<sqrt6`
उत्तर
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
∴ `tan^-1 x/2+tan^-1 x/3=pi/4,`
⇒ `tan^-1((x/2+x/3)/(1-x/2xxx/3))=pi/4`
⇒ `tan^-1(((5x)/6)/((6-x^2)/6))=pi/4`
⇒ `(5x)/(6-x^2)=tan pi/4`
⇒ `(5x)/(6-x^2)=1`
⇒ `5x=6-x^2`
⇒ `x^2+5x-6=0`
⇒ `(x-1) (x+6)=0`
⇒ x = 1 `[because0<x<sqrt6]`
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