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Question

i) Solve for x:tan1(x1)+tan1x+tan1(x+1)=tan13x.
ii) Prove that tan1(6x8x3112x2)tan1(4x14x2)=tan12x;|2x|<13

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Solution

(i) tan1(x1)+tan1x+tan1(x+1)=tan13x

tan1(x1)+tan1(x+1)=tan13xtan1x

tan1(x1+x+11(x1)(x+1))=tan1(3xx1+3x2)

2x1(x21)=2x1+3x2

x(1+3x2)=x(2x2)

x(1+3x22+x2)=0

x(4x21)=0

x=0;4x21=0

x2=14

x=±12

x=0,±12

(ii) Take LHS,

tan1(6x8x3112x2)tan1(4x14x2)

tan1⎜ ⎜ ⎜ ⎜6x8x3112x24x14x21+6x8x3112x2×4x14x2⎟ ⎟ ⎟ ⎟

tan1((6x8x3)(14x2)4x(112x2)(112x2)(14x2)+(6x8x3)4x)

tan1(6x24x38x3+32x54x+48x314x212x2+48x4+24x232x4)

tan1(32x5+16x3+2x16x4+8x2+1)

tan1(2x(16x4+8x+1)16x4+8x+1)

tan12x= RHS

Hence proved.


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