# Transform cot B + tan B to csc B sec B

Transform $\mathrm{cot}B+\mathrm{tan}B\to \mathrm{csc}B\mathrm{sec}B$
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gwibdaithq
Use the quotient identities: $\mathrm{cot}B+\mathrm{tan}B=\left(\frac{\mathrm{cos}B}{\mathrm{sin}B}\right)+\left(\frac{\mathrm{sin}B}{\mathrm{cos}B}\right)$
Add using the LCD which is $\mathrm{sin}B\mathrm{cos}B$: $\mathrm{cot}B+\mathrm{tan}B=\left(\frac{\mathrm{cos}B}{\mathrm{sin}B}\right)\left(\frac{\mathrm{cos}B}{\mathrm{cos}B}\right)+\left(\frac{\mathrm{sin}B}{\mathrm{cos}B}\right)\left(\mathrm{sin}B+\mathrm{sin}B\right)$
$\mathrm{cot}B+\mathrm{tan}B=\frac{{\mathrm{cos}}^{2}B}{\mathrm{sin}B\mathrm{cos}B}+\frac{{\mathrm{sin}}^{2}B}{\mathrm{sin}B\mathrm{cos}B}$
$\mathrm{cot}B+\mathrm{tan}B=\frac{{\mathrm{cos}}^{2}B+{\mathrm{sin}}^{2}B}{\mathrm{sin}B}\mathrm{cos}B$
Use the Pythagorean identity: ${\mathrm{cos}}^{2}B+{\mathrm{sin}}^{2}B=1$
$\mathrm{cot}B+\mathrm{tan}B=\frac{1}{\mathrm{sin}B}\mathrm{cos}B$
$\mathrm{cot}B+\mathrm{tan}B=\left(\frac{1}{\mathrm{sin}B}\right)\left(\frac{1}{\mathrm{cos}B}\right)$
Use reciprocal identities:
$\mathrm{cot}B+\mathrm{tan}B=\mathrm{csc}B\mathrm{sec}B$
Jeffrey Jordon