If tan−1a−11=tan−1x1+tan−1a2−x+11,thenxis

Given,tan−1a−11=tan−1x1+tan−1a2−x+11or, tan−1a−11=tan−11−x(a2−x+1)1x1+a2−x+11or, tan−1a−11=tan−1a2x−x2+x−1a2+1or, (a−1)(a2+1)=a2x−x2+x−1or, a3+a−a2−1=a2x−x2+x−1or, a2(a−x)−(a2−x2)+(a−x)=0or, (a−x)(a2−a−x+1)=0or, x=a,a2−a+1.

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If { ^{ - 1}}{1}{{a - 1}} = { ^{ - 1}}{1}{x} + { ^{ - 1}}{1}{{

Question

If $tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{1}{x} + tan^{- 1} \frac{1}{a ^{2} - x + 1}, t h e n x i s$

a

a^{3}

a^{2} - a + 1

a^{2} + a - 1

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tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{1}{x} + tan^{- 1} \frac{1}{a ^{2} - x + 1}, t h e n x i s

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Text solutionVerified

Given,

tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{1}{x} + tan^{- 1} \frac{1}{a ^{2} - x + 1}

or,

tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{\frac{1}{x} + \frac{1}{a ^{2} - x + 1}}{1 - \frac{1}{x ( a ^{2} - x + 1 )}}

or,

tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{a ^{2} + 1}{a ^{2} x - x ^{2} + x - 1}

or,

(a - 1) (a^{2} + 1) = a^{2} x - x^{2} + x - 1

or,

a^{3} + a - a^{2} - 1 = a^{2} x - x^{2} + x - 1

or,

a^{2} (a - x) - (a^{2} - x^{2}) + (a - x) = 0

or,

(a - x) (a^{2} - a - x + 1) = 0

or,

x = a, a^{2} - a + 1

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Question Text	If $tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{1}{x} + tan^{- 1} \frac{1}{a ^{2} - x + 1}, t h e n x i s$
Answer Type	Text solution:1
Upvotes	150

If tan−1a−11​=tan−1x1​+tan−1a2−x+11​,thenxis

Text solutionVerified

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If $tan^{- 1} \frac{1}{a - 1} = tan^{- 1} \frac{1}{x} + tan^{- 1} \frac{1}{a ^{2} - x + 1}, t h e n x i s$