SOLUTION: Pove that 1+sin(theta)divided by 1-Sin(theta)- 1-Sin(theta) Divided by 1+Sin(theta)= 4Tan(theta)times Sec(theta) Show step by step how to change the left side of the equation to

Algebra ->  Trigonometry-basics -> SOLUTION: Pove that 1+sin(theta)divided by 1-Sin(theta)- 1-Sin(theta) Divided by 1+Sin(theta)= 4Tan(theta)times Sec(theta) Show step by step how to change the left side of the equation to      Log On


   

Question 109302This question is from textbook Algebra and Trigonomometry
: Pove that
1+sin(theta)divided by 1-Sin(theta)- 1-Sin(theta) Divided by 1+Sin(theta)=
4Tan(theta)times Sec(theta)
Show step by step how to change the left side of the equation to equal the right side of the equation. This question is from textbook Algebra and Trigonomometry

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let's only work on the left side of the equation. So we are working on:
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where x represents theta ...
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Multiply the first of these two terms by %281%2Bsin%28x%29%29%2F%281%2Bsin%28x%29%29 and multiply the second
of these two terms by %281-sin%28x%29%29%2F%281-sin%28x%29%29. Note that in each of these multipliers
you are actually multiplying by 1 because the numerator equals the denominator. Therefore,
in effect you are not changing the two terms. We have now converted the problem to:
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Notice the products of the denominators ... in both terms the denominators multiply out
to give 1 - sin^2(x). And if you recall the identity that sin^2(x) + cos^2(x) = 1, then
you can see that cos^2(x) = 1 - sin^2(x). Therefore we can replace the denominators
in the two terms with cos^2(x) which we split up into cos(x)*cos(x) and the problem then
reduces to:
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Now let's multiply out the numerators in the two terms. The numerator of the first term
is the product of %281%2Bsin%28x%29%29%2A%281%2Bsin%28x%29%29 and this multiplies out to:
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1+%2B+2sin%28x%29+%2B+sin%5E2%28x%29
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and the numerator of the second term is the product of %281+-+sin%28x%29%29%2A%281-sin%28x%29%29
and this multiplies out to:
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1+-2sin%28x%29%2B+sin%5E2%28x%29
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These two results make the our expression become:
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Note that the denominators of the 2 terms are the same, so the numerators can be combined over
the common denominator. This leads to:
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When you remove the parentheses around the second half of the numerator, the minus sign
preceding those parentheses causes you to change the signs of all the terms inside the
parentheses and you get:
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Now in the numerator the 1 and the -1 cancel out and the sin%5E2%28x%29 and the -sin%5E2%28x%29
also cancel out. You are therefore left with:
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%282sin%28x%29+%2B+2+sin%28x%29%29%2F%28cos%28x%29%2Acos%28x%29%29
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And the terms in the numerator combine to give:
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%284sin%28x%29%29%2F%28cos%28x%29%2Acos%28x%29%29
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and we can separate this as follows:
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4%2A%28sin%28x%29%2Fcos%28x%29%29%2A%281%2Fcos%28x%29%29
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but by trig identities we know that sin%28x%29%2Fcos%28x%29+=+tan%28x%29 and 1%2Fcos%28x%29+=+sec%28x%29
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Making these substitutions, we end up with:
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4%2A%28sin%28x%29%2Fcos%28x%29%29%2A%281%2Fcos%28x%29%29=+4%2Atan%28x%29%2Asec%28x%29
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And if you look back at the original problem, you find that this is the same as the right
side of the identity you were to prove by showing that the left side could be converted to
the right side ... which could be done as shown above.
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Hope this is understandable and helps you to see your way through the problem.
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