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BabaBlast [244]
3 years ago
5

How would you prove this trigonomic identity? Please show steps.

Mathematics
1 answer:
Agata [3.3K]3 years ago
4 0

Start by combining the fractions:

\dfrac{\cos\alpha}{1+\sin\alpha}\cdot\dfrac{1-\sin\alpha}{1-\sin\alpha}+\dfrac{\cos\alpha}{1-\sin\alpha}\cdot\dfrac{1+\sin\alpha}{1+\sin\alpha}

\dfrac{\cos\alpha(1-\sin\alpha)+\cos\alpha(1+\sin\alpha)}{(1+\sin\alpha)(1-\sin\alpha)}

\dfrac{2\cos\alpha}{1-\sin^2\alpha}

Recall the Pythagorean identity:

\dfrac{2\cos\alpha}{\cos^2\alpha}

Then cancel a factor of \cos\alpha and use the definition of secant:

\dfrac2{\cos\alpha}=\boxed{2\sec\alpha}

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F(x+1)+f(x+2)=2x+3; f(x)=?
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Answer:

F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1

Step-by-step explanation:

1. \mathrm{Subtract\:}f\left(x+2\right)\mathrm{\:from\:both\:sides}\\
F\left(x+1\right)+f\left(x+2\right)-f\left(x+2\right)=2x+3f-f\left(x+2\right)



2. \mathrm{Simplify}\\
F\left(x+1\right)=-xf+f+2x\\


3. \mathrm{Divide\:both\:sides\:by\:}x+1;\quad \:x\ne \:-1\\
\frac{F\left(x+1\right)}{x+1}=-\frac{xf}{x+1}+\frac{f}{x+1}+\frac{2x}{x+1};\quad \:x\ne \:-1

4. \mathrm{Simplify}\\
F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1

Final Answer: F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1

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