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kati45 [8]
3 years ago
13

Verify cos(x+y)/cos(x-y) = cot(x-y)/cot(x+y)

Mathematics
1 answer:
JulijaS [17]3 years ago
3 0

\bf \cfrac{cos(x+y)}{cos(x-y)}=\cfrac{cot(x-y)}{cot(x+y)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{expanding the left-hand-side}}{\cfrac{cos(x)cos(y)-sin(x)sin(y)}{cos(x)cos(y)+sin(x)sin(y)}} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{expanding the right-hand-side}}{\cfrac{~~ \frac{cos(x)cos(y)+sin(x)sin(y)}{sin(x)cos(y)-cos(x)sin(y)}~~}{\frac{cos(x)cos(y)-sin(x)sin(y)}{sin(x)cos(y)+cos(x)sin(y)}}} \\\\\\ \cfrac{cos(x)cos(y)+sin(x)sin(y)}{sin(x)cos(y)-cos(x)sin(y)}\cdot \cfrac{sin(x)cos(y)+cos(x)sin(y)}{cos(x)cos(y)-sin(x)sin(y)}

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3x + 1 < −2x + 8
Fantom [35]

Hello there!


The correct answer is B. Why? Well, we cannot choose A, because we don't have a multiplication sign. We cannot choose C because we also don't have a multiplication sign. However, since we have a PLUS sign and a Subtraction sign, we can say that the correct answer is option B.


3x + 1 < -2x + 8

We wanna start by subtracting 1 on both sides

3x + 1 - 1 < -2x + 8 - 1

3x < -2x + 7

Add 2x on both sides

3x + 2x < -2x + 7 + 2x

5x < 7

x < 7/5


I hope this answer helps! As always, it is my pleasure to help students like you. If you have any additional questions, feel free to comment them down below. I'll see you around!

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