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प्रश्न
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
उत्तर
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
`therefore tan^-1 2x+tan^-1 3x=npi+(3pi)/4`
⇒ `tan^-1((2x+3x)/(1-2x xx3x))=npi+(3pi)/4`
⇒ `(5x)/(1-6x^2)=tan(npi+(3pi)/4)`
⇒ `(5x)/(1-6x^2)=-1`
⇒ `5x=-1+6x^2`
⇒ `6x^2-5x-1=0`
⇒ `(6x+1)(x-1)=0`
⇒ `x=-1/6` [As x=1 is not satisfying the equation]
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