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BloomLocke367
 one year ago
Verify the identity
BloomLocke367
 one year ago
Verify the identity

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BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large\tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}\)

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2\[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\] Change everything to sines and cosines \[\large \frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\] Now we need a common denominator...what would that be?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(\sin\theta+\cos\theta\)?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Not quite....if we multiply the 2 denominators we get \(\large sin\theta cos\theta\) that that would be our common denominator...so what we need to do is \[\large \frac{sin\theta}{sin\theta} \times \frac{sin\theta}{cos\theta} + \frac{cos\theta}{cos\theta} \times \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\] Simplify that down a bit \[\large \frac{sin^2 \theta}{sin\theta cos\theta} + \frac{cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] Now we can combine those 2 over the common denominator \[\large \frac{sin^2 \theta + cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos \theta}\] now what else do we have to do?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0multiply both sides by sinthetacostheta?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Not nearly as difficult...what is the mother of all trigonometric identities that we should know? \[\large sin^2 \theta + cos^2 \theta = 1\]

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0is that what you get when you multiply both sides by the denominators?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Nope...so what we had was \[\large \frac{sin^2 \theta + cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] On the left one...we see we have \(\large sin^2 \theta + cos^2 \theta\) which we now know equals 1...so now we have \[\large \frac{1}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] and the identity is confirmed!

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Let me gather everything together...maybe it will help you see it \[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\] Changing everything to sines and cosines \[\large \frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\] Finding the common denominator...which is \(\large sin\theta cos\theta\) ...we need to multiply the first fraction by sin/sin and the second fraction by cos/cos to even things out \[\large \frac{sin\theta}{sin\theta} \times \frac{sin\theta}{cos\theta} + \frac{cos\theta}{cos\theta} \times \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\] Simplify those products \[\large \frac{sin^2\theta}{sin\theta cos\theta} + \frac{cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] Now that the two fractions have a common denominator...we combine them over that \[\large \frac{\color \red{sin^2\theta + cos^2 \theta}}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] Now....we remember the Trigonometric Identity that states \(\large \color\red{sin^2\theta + cos^2 \theta = 1}\) so we use that here... \[\large \frac{1}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\] Which verifies the identity
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