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September 9th, 2015, 07:22 PM   #1
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Prove (secA-tanA+1)/(secA+tanA+1)=(1-sinA)/cosA
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September 9th, 2015, 08:46 PM   #2
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$\displaystyle \dfrac{\sec A-\tan A+1}{\sec A+\tan A+1}=\dfrac{1-\sin A+\cos A}{1+\sin A+\cos A}\cdot\dfrac{1+\sin A-\cos A}{1+\sin A-\cos A}=\dfrac{\cos A}{1+\sin A}=\dfrac{1-\sin A}{\cos A}$
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September 9th, 2015, 10:41 PM   #3
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Alternatively, as 1 = sec²A - tan²A = (secA+tanA)(secA-tanA),
(secA-tanA+1)/(secA+tanA+1) = (secA-tanA+(secA+tanA)(secA-tanA))/(1+secA+tanA)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\,$ = secA - tanA
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\,$ = (1 - sinA)/cosA.
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September 10th, 2015, 01:52 AM   #4
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Quote:
Originally Posted by Supriyo View Post
Prove (secA-tanA+1)/(secA+tanA+1)=(1-sinA)/cosA
$\displaystyle \dfrac{\sec A-\tan A+1}{\sec A+\tan A+1}=\dfrac{1-\sin A}{\cos A}\ \ \ (1)
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\sec A=\dfrac{1}{\cos A}\ \ \ (2)
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\tan A=\dfrac{\sin A}{\cos A}\ \ \ (3)
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(1),\ (2), \ (3)\ \Longrightarrow\ \dfrac{1-\sin A+\cos A}{1+\sin A+\cos A} = \dfrac{1-\sin A}{\cos A}\ \ \ (4)
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(4)\Longleftrightarrow\ \cos A(1-\sin A+\cos A)=(1-\sin A)(1+\sin A+\cos A)
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\Longleftrightarrow\ \cos A- \sin A \cos A +\cos^2A = 1+\sin A+\cos A -\sin A - \sin^2A- \sin A\cos A
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\text{After simplifications, we have:}\ \cos^2 A = 1-\sin^2 A \ \Longleftrightarrow\ \cos^2 A = \cos^2 A \ (\text{True})$

Last edited by skipjack; September 10th, 2015 at 03:50 AM.
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September 10th, 2015, 08:20 AM   #5
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$\displaystyle \color{blue} {\dfrac{1-\sin A}{\cos A}= \dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}= \sec A - \tan A
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Note\ \sec A = s,\ \ \tan A=t.
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We\ know\ that:
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\sec^2A-\tan^2A = 1\ \Longrightarrow\ s^2-t^2=1\ (*)
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Equality\ \ becomes:
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\dfrac{s-t+1}{s+t+1}=s-t\Longleftrightarrow s-t+1=(s-t)(s+t+1)\Longleftrightarrow
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\Longleftrightarrow s-t+1=(s-t)[(s+t)+1] \Longleftrightarrow s-t+1= s^2-t^2+s-t \stackrel{(*)}{\Longleftrightarrow}
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\Longleftrightarrow \ s-t+1 = 1+s-t \Longleftrightarrow \ s-t+1 = s-t+1 \ \ (true)}$
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