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प्रश्न
Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`
उत्तर
Consider the left hand side
L.H.S = `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))`
We know that,
`tan^(-1)(A)-tan^(-1)(B)= tan^(-1)((A-B)/(1+AB))`
Thus, L.H.S = `tan^(-1)(((6x-8x^3)/(1-12x^2)-(4x)/(1-4x^2))/(1+((6x-8x^3)/(1-12x^2))((4x)/(1-4x^2))))`
`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/(((1-12x^2)(1-4x^2))/(1+(4x(6x-8x^3))/((1-12x^2)(1-4x^2)))))`
`=tan^(-1)((((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)))/(((1-12x^2)(1-4x^2)+4x(6x-8x^3))/((1-12x^2)(1-4x^2))))`
`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)+4x(6x-8x^3)))`
`=tan^(-1)((6x-24x^3-8x^3+32x^5-4x+48x^3)/(1-4x^2-12x^2+48x^4+24x^2-32x^4))`
`=tan^(-1)((32x^5+16x^3+2x)/(16x^4+8x^2+1))`
`=tan^(-1)((2x(16x^4+8x^2+1))/(16x^4+8x^2+1))`
= tan-12x
Thus, L.H.S=R.H.S
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