1. ## trig proof

prove that

(1-sinA)/(1-secA)-(1+sinA)/(1+secA)=2cotA(cosA-cosecA)

if anyone can show me how to do this it would be greatly appreciated

2. Originally Posted by thehaanz
prove that

(1-sinA)/(1-secA)-(1+sinA)/(1+secA)=2cotA(cosA-cosecA)

if anyone can show me how to do this it would be greatly appreciated
$\frac{1 - \sin{A}}{1 - \sec{A}} - \frac{1 + \sin{A}}{1 + \sec{A}} = \frac{1 - \sin{A}}{1 - \frac{1}{\cos{A}}} - \frac{1 + \sin{A}}{1 + \frac{1}{\cos{A}}}$

$= \frac{1 - \sin{A}}{\frac{\cos{A} - 1}{\cos{A}}} - \frac{1 + \sin{A}}{\frac{\cos{A} + 1}{\cos{A}}}$

$= \frac{\cos{A}(1 - \sin{A})}{\cos{A} - 1} - \frac{\cos{A}(1 + \sin{A})}{\cos{A} + 1}$

$= \frac{\cos{A}(\cos{A} + 1)(1 - \sin{A})}{(\cos{A} - 1)(\cos{A} + 1)} - \frac{\cos{A}(\cos{A} - 1)(1 + \sin{A})}{(\cos{A} - 1)(\cos{A} + 1)}$

$= \frac{\cos{A}(\cos{A} + 1)(1 - \sin{A}) - \cos{A}(\cos{A} - 1)(1 + \sin{A})}{\cos^2{A} - 1}$

$= \frac{\cos{A}(\cos{A} - 1)(1 + \sin{A}) - \cos{A}(\cos{A} + 1)(1 - \sin{A})}{\sin^2{A}}$

$= \frac{\cos{A}[(\cos{A} - 1)(1 + \sin{A}) - (\cos{A} + 1)(1 - \sin{A})]}{\sin^2{A}}$

$= \frac{\cos{A}(\cos{A} + \cos{A}\sin{A} - 1 - \sin{A} - \cos{A} + \cos{A}\sin{A} - 1 + \sin{A})}{\sin^2{A}}$

$= \frac{\cos{A}(2\cos{A}\sin{A} - 2)}{\sin^2{A}}$

$= \frac{2\cos{A}(\cos{A}\sin{A} - 1)}{\sin^2{A}}$

$= 2\cot{A}\left(\frac{\cos{A}\sin{A} - 1}{\sin{A}}\right)$

$= 2\cot{A}\left(\frac{\cos{A}\sin{A}}{\sin{A}} - \frac{1}{\sin{A}}\right)$

$= 2\cot{A}(\cos{A} - \csc{A})$.

3. Hello, thehaanz!

Prove that: . $\frac{1-\sin A}{1-\sec A} -\frac{1+\sin A}{1+\sec A} \;=\;2\cot A\,(\cos A- \csc A)$

$\frac{1-\sin A}{1-\sec A} - \frac{1+\sin A}{1 + \sec A} \;\;=\;\;\frac{(1-\sin A)(1+\sec A) - (1+\sin A)(1-\sec A)}{(1-\sec A)(1 + \sec A)}$

. . . . . . . . . . . . . . $=\;\;\frac{1 + \sec A - \sin A - \sin A\sec A - 1 + \sec A - \sin A + \sin A\sec A}{1-\sec^2A}$

. . . . . . . . . . . . . . $=\;\;\frac{2\sec A - 2\sin A}{-(\sec^2\!A - 1)} \;\;=\;\;\frac{-2(\sin A - \sec A)}{-\tan^2\!A} \;\;=\;\;\frac{2(\sin A - \sec A)}{\tan^2\!A}$

. . . . . . . . . . . . . . $=\;\;\frac{2\left(\sin A - \dfrac{1}{\cos A}\right)}{\dfrac{\sin^2\!A}{\cos^2\!A}} \;\;=\;\;2\cdot\frac{\cos^2\!A}{\sin^2\!A}\left(\s in A - \frac{1}{\cos A}\right)$

. . . . . . . . . . . . . . $=\;\;2\cdot\frac{\cos A}{\sin A}\cdot\underbrace{\frac{\cos A}{\sin A}\left(\sin A - \frac{1}{\cos A}\right)}$

. . . . . . . . . . . . . . $=\quad 2\cdot\frac{\cos A}{\sin A}\cdot\left(\cos A - \frac{1}{\sin A}\right)$

. . . . . . . . . . . . . . $=\qquad 2\cdot\cot A(\cos A - \csc A)$

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# 2cotA(cosA-cosecA

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