Prove that \(\displaystyle{\tan{{\left[{\frac{{\pi}}{{{4}}}}+{\frac{{{1}}}{{{2}}}}{{\cos}^{{-{1}}}{\left({\frac{{{a}}}{{{b}}}}\right)}}\right]}}}+{\tan{{\left[{\frac{{\pi}}{{{4}}}}-{\frac{{{1}}}{{{2}}}}{{\cos}^{{-{1}}}{\left({\frac{{{a}}}{{{b}}}}\right)}}\right]}}}={\frac{{{2}{b}}}{{{a}}}}\)

Landen Barber

Landen Barber

Answered question

2022-03-29

Prove that tan[π4+12cos1(ab)]+tan[π412cos1(ab)]=2ba

Answer & Explanation

Malia Booth

Malia Booth

Beginner2022-03-30Added 16 answers

Using
tan(A+B)=tanA+tanB1tanAtanB
There are
tan[π4±12cos1ab]=1±tan12cos1ab1tan12cos1ab
Hence
tan[π4+12cos1(ab)]+tan[π412cos1(ab)]=21+tan2(12cos1ab)1tan2(12cos1ab)
using the half-angle formula, proceed.
cosα=1tan2α21+tan2α2
There are
tan[π4+12cos1(ab)]+tan[π412cos1(ab)]=2coscos1ab=2ba

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