BloomLocke367
  • BloomLocke367
Verify the identity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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BloomLocke367
  • BloomLocke367
\(\Large\tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}\)
BloomLocke367
  • BloomLocke367
@ganeshie8
johnweldon1993
  • johnweldon1993
\[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\] 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?

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BloomLocke367
  • BloomLocke367
\(\sin\theta+\cos\theta\)?
johnweldon1993
  • johnweldon1993
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?
BloomLocke367
  • BloomLocke367
multiply both sides by sinthetacostheta?
johnweldon1993
  • johnweldon1993
Not nearly as difficult...what is the mother of all trigonometric identities that we should know? \[\large sin^2 \theta + cos^2 \theta = 1\]
BloomLocke367
  • BloomLocke367
is that what you get when you multiply both sides by the denominators?
johnweldon1993
  • johnweldon1993
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!
BloomLocke367
  • BloomLocke367
what?
johnweldon1993
  • johnweldon1993
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
BloomLocke367
  • BloomLocke367
ohhh okay

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