Prove the identity:

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\mathrm{csc}x+\mathrm{cot}x$

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\mathrm{csc}x+\mathrm{cot}x$

Dayanara Terry
2022-07-15
Answered

Prove the identity:

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\mathrm{csc}x+\mathrm{cot}x$

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\mathrm{csc}x+\mathrm{cot}x$

You can still ask an expert for help

treccinair

Answered 2022-07-16
Author has **18** answers

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}=\mathrm{csc}x+\mathrm{cot}x$

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}\ast \frac{(1+\mathrm{cos}x)}{(1+\mathrm{cos}x)}$

$=\frac{\mathrm{sin}x-\mathrm{sin}x\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}\text{}\text{}\text{}{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$

$=\frac{\mathrm{sin}x+\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x}\text{}\text{}\text{}{\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$

$=\frac{\mathrm{sin}x}{{\mathrm{sin}}^{2}x}+\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x}$

$=\frac{1}{\mathrm{sin}x}+\frac{\mathrm{cos}x}{\mathrm{sin}x}$

$=\mathrm{csc}x-\mathrm{cot}x$

$\frac{\mathrm{sin}x}{1-\mathrm{cos}x}\ast \frac{(1+\mathrm{cos}x)}{(1+\mathrm{cos}x)}$

$=\frac{\mathrm{sin}x-\mathrm{sin}x\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}\text{}\text{}\text{}{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$

$=\frac{\mathrm{sin}x+\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x}\text{}\text{}\text{}{\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$

$=\frac{\mathrm{sin}x}{{\mathrm{sin}}^{2}x}+\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x}$

$=\frac{1}{\mathrm{sin}x}+\frac{\mathrm{cos}x}{\mathrm{sin}x}$

$=\mathrm{csc}x-\mathrm{cot}x$

Jeffrey Jordon

Answered 2022-08-01
Author has **2581** answers

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