CMIIh

2020-11-23

Solve the equation $\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }$

gotovub

We work using the left side. Write in terms of sine and cosine:
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{1}{\frac{1}{\mathrm{sin}\theta }-\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }}$
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{1}{\frac{1-\mathrm{cos}\theta }{\mathrm{sin}\theta }}$
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{\mathrm{sin}\theta }{1-\mathrm{cos}\theta }$
Multiply by $\frac{1+\mathrm{cos}\theta }{1+\mathrm{cos}\theta }$:
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{\mathrm{sin}}{1-\mathrm{cos}\theta }×\frac{1+\mathrm{cos}\theta }{1+\mathrm{cos}\theta }$
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}{1-{\mathrm{cos}}^{2}\theta }$
Use the Pythagorean identity:
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}{{\mathrm{sin}}^{2}\theta }$
$\frac{1}{\mathrm{csc}\theta -\mathrm{cos}\theta }=\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }$

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