Rewrite the expression in terms of the given function.

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x$$

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x$$

Kymani Hatfield
2022-10-22
Answered

Rewrite the expression in terms of the given function.

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x$$

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x$$

You can still ask an expert for help

honejata1

Answered 2022-10-23
Author has **10** answers

Answer:

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x\phantom{\rule{0ex}{0ex}}=\frac{1-\mathrm{cos}x+(1+\mathrm{cos}x)(\mathrm{cos}x)}{1-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1+{\mathrm{cos}}^{2}x}{1-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1+{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1}{{\mathrm{sin}}^{2}x}+\frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{0ex}{0ex}}={\mathrm{csc}}^{2}x+{\mathrm{cot}}^{2}x\phantom{\rule{0ex}{0ex}}=(1+{\mathrm{cot}}^{2}x)+\mathrm{cot}x\phantom{\rule{0ex}{0ex}}=1+2{\mathrm{cot}}^{2}x$$

$$\frac{1}{1+\mathrm{cos}x}+\frac{\mathrm{cos}x}{1-\mathrm{cos}x};\mathrm{cot}x\phantom{\rule{0ex}{0ex}}=\frac{1-\mathrm{cos}x+(1+\mathrm{cos}x)(\mathrm{cos}x)}{1-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1+{\mathrm{cos}}^{2}x}{1-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1+{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{1}{{\mathrm{sin}}^{2}x}+\frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{0ex}{0ex}}={\mathrm{csc}}^{2}x+{\mathrm{cot}}^{2}x\phantom{\rule{0ex}{0ex}}=(1+{\mathrm{cot}}^{2}x)+\mathrm{cot}x\phantom{\rule{0ex}{0ex}}=1+2{\mathrm{cot}}^{2}x$$

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