$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x=\mathrm{sin}x$

Ismael Molina
2022-07-16
Answered

Prove the identity

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x=\mathrm{sin}x$

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x=\mathrm{sin}x$

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Bianca Chung

Answered 2022-07-17
Author has **16** answers

Prove the identity

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x=\mathrm{sin}x$

Proof:

L.H.S.

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}x}-\frac{\mathrm{cos}x}{\mathrm{sin}x}\cdot \mathrm{cos}x\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{2}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}x\text{}\text{}\text{}R.H.S$

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x=\mathrm{sin}x$

Proof:

L.H.S.

$\mathrm{csc}x-\mathrm{cot}x\cdot \mathrm{cos}x\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}x}-\frac{\mathrm{cos}x}{\mathrm{sin}x}\cdot \mathrm{cos}x\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{2}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}x\text{}\text{}\text{}R.H.S$

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Please explain at beginner level.

${\int}_{0}^{1}\sqrt{{x}^{2}-2x+1}dx$

so I simplified it algebraically to

${\int}_{0}^{1}\sqrt{(x-1{)}^{2}}dx$

which of course is

${\int}_{0}^{1}|x-1|dx$

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Please explain at beginner level.

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