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BloomLocke367

  • one year ago

Verify the identity

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  1. BloomLocke367
    • one year ago
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    \(\Large\tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}\)

  2. BloomLocke367
    • one year ago
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    @ganeshie8

  3. johnweldon1993
    • one year ago
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    \[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\] 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?

  4. BloomLocke367
    • one year ago
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    \(\sin\theta+\cos\theta\)?

  5. johnweldon1993
    • one year ago
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    Not 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?

  6. BloomLocke367
    • one year ago
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    multiply both sides by sinthetacostheta?

  7. johnweldon1993
    • one year ago
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    Not nearly as difficult...what is the mother of all trigonometric identities that we should know? \[\large sin^2 \theta + cos^2 \theta = 1\]

  8. BloomLocke367
    • one year ago
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    is that what you get when you multiply both sides by the denominators?

  9. johnweldon1993
    • one year ago
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    Nope...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!

  10. BloomLocke367
    • one year ago
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    what?

  11. johnweldon1993
    • one year ago
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    Let 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

  12. BloomLocke367
    • one year ago
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    ohhh okay

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