Transform $\mathrm{cot}B+\mathrm{tan}B\to \mathrm{csc}B\mathrm{sec}B$

ossidianaZ
2021-05-13
Answered

Transform $\mathrm{cot}B+\mathrm{tan}B\to \mathrm{csc}B\mathrm{sec}B$

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gwibdaithq

Answered 2021-05-14
Author has **84** answers

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$\mathrm{sin}B\mathrm{cos}B$ :
$\mathrm{cot}B+\mathrm{tan}B=\left(\frac{\mathrm{cos}B}{\mathrm{sin}B}\right)\left(\frac{\mathrm{cos}B}{\mathrm{cos}B}\right)+\left(\frac{\mathrm{sin}B}{\mathrm{cos}B}\right)(\mathrm{sin}B+\mathrm{sin}B)$

$\mathrm{cot}B+\mathrm{tan}B=\frac{{\mathrm{cos}}^{2}B}{\mathrm{sin}B\mathrm{cos}B}+\frac{{\mathrm{sin}}^{2}B}{\mathrm{sin}B\mathrm{cos}B}$

$\mathrm{cot}B+\mathrm{tan}B=\frac{{\mathrm{cos}}^{2}B+{\mathrm{sin}}^{2}B}{\mathrm{sin}B}\mathrm{cos}B$

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:

$\mathrm{cot}B+\mathrm{tan}B=\mathrm{csc}B\mathrm{sec}B$

Add using the LCD which is

Use the Pythagorean identity:

Use reciprocal identities:

Jeffrey Jordon

Answered 2021-08-11
Author has **2027** answers

Transformation:

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