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प्रश्न
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
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उत्तर
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
∴ tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`
⇒ `tan^-1((x+2+x-2)/(1-(x+2)xx(x-2)))=tan^-1 8/79`
⇒ `(2x)/(1-x^2+4)=8/79`
⇒ `x/(5-x^2)=4/79`
⇒ `79x=20-4x^2`
⇒ `4x^2+79x-20=0`
⇒ `4x^2+80x-x-20=0`
⇒ `(4x-1)(x+20)=0`
⇒ `x=1/4 or - 20`
∴ `x=1/4` `[becausex>0]`
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