Proving the identity $\mathrm{csc}x-\mathrm{sin}x=(\mathrm{cot}x)(\mathrm{cos}x)$

Ainsley Zimmerman
2022-05-08
Answered

Proving the identity $\mathrm{csc}x-\mathrm{sin}x=(\mathrm{cot}x)(\mathrm{cos}x)$

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gudstrufy47j

Answered 2022-05-09
Author has **16** answers

So starting with LHS:

$\begin{array}{rl}\frac{1}{\mathrm{sin}x}-\mathrm{sin}x& =\frac{1-(\mathrm{sin}x{)}^{2}}{\mathrm{sin}x}\\ & =\frac{(\mathrm{cos}x{)}^{2}}{\mathrm{sin}x}\end{array}$

$\frac{\mathrm{cos}x.\mathrm{cos}x}{\mathrm{sin}x}$

$=\frac{\mathrm{cos}x}{\mathrm{sin}x}.\mathrm{cos}x$

$=\mathrm{cot}x.\mathrm{cos}x$

$\begin{array}{rl}\frac{1}{\mathrm{sin}x}-\mathrm{sin}x& =\frac{1-(\mathrm{sin}x{)}^{2}}{\mathrm{sin}x}\\ & =\frac{(\mathrm{cos}x{)}^{2}}{\mathrm{sin}x}\end{array}$

$\frac{\mathrm{cos}x.\mathrm{cos}x}{\mathrm{sin}x}$

$=\frac{\mathrm{cos}x}{\mathrm{sin}x}.\mathrm{cos}x$

$=\mathrm{cot}x.\mathrm{cos}x$

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I tried the following substitution :

I'm really disturbed by the