How is it that \ln(2+\tan\theta) - \frac{1}{2}\ln(1+\tan^2\theta) = \ln (\frac{2 +

compyac

compyac

Answered question

2022-04-21

How is it that
ln(2+tanθ)12ln(1+tan2θ)=ln(2+tanθsecθ)

Answer & Explanation

August Moore

August Moore

Beginner2022-04-22Added 17 answers

ln(2+tanθ)12ln(1+tan2θ)=ln(2+tanθsecθ)
=ln(2+tanθ)ln(secθ)
=ln(2+tanθsecθ)
Klanglinkmgk

Klanglinkmgk

Beginner2022-04-23Added 13 answers

ln(1+tanθ)12ln(1+tan2θ)=ln(2+tanθsecθ)
=ln((1+tanθ)1+tan2θ)
=ln((1+tanθ)sec2θ)
=ln(1+tanθ|secθ|)

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