What is $tan (arcsin (\frac{12}{13}))$ $tan(arcsin(12/13))$ ?

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Answer 1 · 2015-07-21T21:32:44

$tan (arcsin (\frac{12}{13})) = \frac{12}{5}$ $tan(arcsin(12/13))=12/5$

Let $θ = arcsin (\frac{12}{13})$ $" "theta=arcsin(12/13)$

This means that we are now looking for $tan θ!$ $color(red)tantheta!$

$\Rightarrow sin (θ) = \frac{12}{13}$ $=>sin(theta)=12/13$

Use the identity,

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

$\Rightarrow \frac{{cos}^{2} θ + {sin}^{2} θ}{{cos}^{2} θ} = \frac{1}{{cos}^{2} θ}$ $=>(cos^2theta+sin^2theta)/cos^2theta=1/cos^2theta$

$\Rightarrow 1 + \frac{{sin}^{2} θ}{{cos}^{2} θ} = \frac{1}{{cos}^{2} θ}$ $=>1+sin^2theta/cos^2theta=1/cos^2theta$

$\Rightarrow 1 + {tan}^{2} θ = \frac{1}{{cos}^{2} θ}$ $=>1+tan^2theta=1/cos^2theta$

$\Rightarrow tan θ = \sqrt{\frac{1}{{cos}^{2} (θ)} - 1}$ $=> tantheta=sqrt(1/cos^2(theta)-1)$

Recall : ${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta= 1-sin^2theta$

$\Rightarrow tan θ = \sqrt{\frac{1}{1 - {sin}^{2} θ} - 1}$ $=>tantheta=sqrt(1/(1-sin^2theta)-1)$

$\Rightarrow tan θ = \sqrt{\frac{1}{1 - {(\frac{12}{13})}^{2}} - 1}$ $=>tantheta=sqrt(1/(1-(12/13)^2)-1)$

$\Rightarrow tan θ = \sqrt{\frac{169}{169 - 144} - 1}$ $=>tantheta=sqrt(169/(169-144)-1$

$\Rightarrow tan θ = \sqrt{\frac{169}{25} - 1}$ $=>tantheta=sqrt(169/25-1)$

$\Rightarrow tan θ = \sqrt{\frac{144}{5}} = \frac{12}{5}$ $=>tantheta=sqrt(144/5)= 12/5$

REMEMBER what we called $θ$ $theta$ was actually $arcsin (\frac{12}{13})$ $arcsin(12/13)$

$\Rightarrow tan (arcsin (\frac{12}{13})) = \frac{12}{5}$ $=>tan(arcsin(12/13))= color(blue)(12/5)$

What is tan(arcsin(12/13)) tan(arcsin(1213))tan(arcsin(12/13)) ?