Walter Clyburn

2022-01-14

If $\frac{a+b}{\mathrm{csc}x}=\frac{a-b}{\mathrm{cot}x}$ show that $\mathrm{csc}x\mathrm{cot}x=\frac{{a}^{2}-{b}^{2}}{4ab}$

sirpsta3u

Suppose $\mathrm{sin}x\ne 0$. Then we have that $\mathrm{csc}x=\frac{1}{\mathrm{sin}x}$ and $\frac{\mathrm{cos}x}{\mathrm{sin}x}=\mathrm{cot}x$
This implies in $\mathrm{cos}x=\frac{a-b}{a+b}$. Therefore, $\mathrm{csc}x\cdot \mathrm{cot}x=\frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}=\frac{\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}$
Notice that $1-{\mathrm{cos}}^{2}x=1-{\left(\frac{a-b}{a+b}\right)}^{2}=\frac{4ab}{{\left(a+b\right)}^{2}}$
Then, $\frac{\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}=\frac{\left(a-b\right)\left(a+b\right)}{4ab}=\frac{{a}^{2}-{b}^{2}}{4ab}$

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