"Finding the value of sin−11213+cos−145+tan−16313" was answered by our community

Alan Smith

Answered question

2022-01-17

Finding the value of

\sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{45}{+} \tan^{- 1} \frac{63}{13}

Answer & Explanation

RizerMix

Expert2022-01-19Added 656 answers

Actually
$\tan^{- 1} \frac{12}{5} + \tan^{- 1} \frac{3}{4} = = π + \tan^{- 1} (\frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \cdot \frac{3}{4}})$
We can notice that
$\frac{π}{2} < \tan^{- 1} \frac{12}{5} + \tan^{- 1} \frac{3}{4} < π$
I am going to prove that
NSK
if $\frac{π}{2} < α + β < \frac{3 π}{2} t h e n α + β = π + \tan^{- 1} (\frac{\tan α + \tan β}{1 - \tan α \cdot \tan β})$
Proof:
$\tan (α + β - π) = \frac{\tan α + \tan (β - π)}{1 - \tan α \tan (β - π)} = \frac{\tan α + \tan β}{1 - \tan α \cdot \tan β}$
As $| α + β - π | < \frac{π}{2}$ and the function tangent is invertible in $[- \frac{π}{2}, \frac{π}{2}]$ it follows that
$α + β - π = \tan^{- 1} (\frac{\tan α + \tan β}{1 - \tan α \cdot \tan β})$ , therefore:
$α + β = π + \tan^{- 1} (\frac{\tan α + \tan β}{1 - \tan α \cdot \tan β})$

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