Question

Write the set of values of x for which tan^{-1}\:\left(\frac{2x}{1+x^2}\right)=2tan^{-1}

Answer 1

Answer:

Step-by-step explanation:

Given  : 2tan−1x=cos−11+x21−x2​

Substitute x=tanθ

⟹2tan−1x=2tan−1(tanθ)2θ

cos−1(1+x21−x2​)

=cos−1(1+tan2θ1−tan2θ​)

=cos−1(cos2θ−sin2θcos2θ−sin2θ​)=cos−1(cos2θ)=2θ=LHS

∴LHS=RHS holds for x=tanθ

∵tanθ∈(0,∞)

⟹x∈(0,∞)

October 15, 2019

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Answer 2

Answer:

Step-by-step explanation:

Given : 2tan-1x=cos-11+x21-x2

Substitute x=tane

2tan-1x=2tan-1(tano)20

cos-1(1+x21-x2)

=cos-1(1+tan 201-tan20)

cos-1(cos20-sin2ecos20-sin2e

)=cos-1(cos20)=28=LHS

:-LHSERHS holds for x=tano

tane E(0,00)

x+(0,…)