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प्रश्न
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
उत्तर
⇒ `cot^-1(x)-cot^-1(x+2)=pi/12`
⇒ `tan^-1(1/x)+cot^-1(1/(x+2))=pi/12` `[because cot^-1x=tan^-1 1/x]`
⇒ `tan^-1((1/x-1/(x+2))/(1+1/(x(x+2))))=pi/12`
⇒ `tan^-1((2/(x(x+2)))/((x^2+2x+1)/(x(x+2))))=pi/12`
⇒ `tan^-1(2/(x^2+2x+1))=pi/12`
⇒ `(2/(x^2+2x+1))=tan pi/12`
⇒ `(2/(x^2+2x+1))=tan(pi/3-pi/4)`
⇒ `(2/(x^2+2x+1))=(tan pi/3-tan pi/4)/(1+tan pi/3xxtan pi/4`
⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)`
⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)xx(sqrt3+1)/(sqrt3+1)`
⇒ `(2/(x^2+2x+1))=2/(sqrt3+1)^2`
⇒ `1/(x+1)^2=1/(sqrt3+1)^2`
⇒ `x+1=sqrt3+1`
⇒ `x=sqrt3`
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