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प्रश्न
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
उत्तर
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
∴ `tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
⇒ `tan^-1(((x-2)/(x-4)+(x+2)/(x+4))/(1-(x-2)/(x-4)xx(x+2)/(x+4)))=pi/4`
⇒ `tan^-1(((x^2+2x-8+x^2-2x-8)/((x-4)(x+4)))/((x^2-16-x^2+4)/((x-4)(x+4))))=pi/4`
⇒ `(2x^2-16)/-12=tan pi/4`
⇒ `(2x^2-16)/-12=1`
⇒ 2x2 - 16 = -12
⇒ 2x2 = 4
⇒ x2 = 2
⇒ `x=+-sqrt2`
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