Question 511681
In this proof you can make use of three identities as follows:
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{{{cos(theta)*sec(theta) = 1}}} which is equivalent to {{{sec(theta) = 1/cos(theta)}}}
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{{{tan(theta) = sin(theta)/cos(theta)}}}
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and
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{{{sin^2(theta)+ cos^2(theta) = 1}}} which is equivalent to {{{cos^2(theta) = 1 - sin^2(theta)}}}
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Start with:
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{{{(sec(theta))(1-sin(theta))(sec(theta)+tan(theta)) = 1}}}
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On the left side replace {{{sec(theta)}}} by its equivalent {{{1/cos(theta)}}} in two places to get:
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{{{(1/cos(theta))(1-sin(theta))(1/cos(theta)+ tan(theta))=1}}}
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Next replace {{{tan(theta)}}} by its equivalent {{{sin(theta)/cos(theta)}}}:
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{{{(1/cos(theta))(1-sin(theta))(1/cos(theta)+ sin(theta)/cos(theta))=1}}}
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Multiply the first two terms in parentheses and the equation becomes:
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{{{((1-sin(theta))/cos(theta))(1/cos(theta)+ sin(theta)/cos(theta))=1}}}
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In the second set of parentheses put everything over the common denominator and this makes the equation become:
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{{{((1-sin(theta))/cos(theta))((1+sin(theta))/cos(theta))=1}}}
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Multiply the two numerators together. Also multiply the two denominators together:
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{{{((1-sin^2(theta))/cos^2(theta))=1}}}
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Now replace the numerator by its equivalent {{{cos^2(theta)}}} and the equation reduces to:
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{{{(cos^2(theta)/cos^2(theta))=1}}}
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Dividing out the left side further reduces the equation to:
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{{{1 = 1}}}
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And since the left side has been reduced until it equaled the right side, the proof has been validated. QED (thus it is demonstrated).
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Hope this helps to familiarize you with some of the trig identities and how to use them.