How do I prove that $\frac{1}{sec A + 1} + \frac{1}{sec A - 1}$ $1/(sec A+1)+1/(sec A-1)$ = $2 csc A cot A$ $2 csc A cot A$ ?

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How do I prove that $\frac{1}{sec A + 1} + \frac{1}{sec A - 1}$ $1/(sec A+1)+1/(sec A-1)$ = $2 csc A cot A$ $2 csc A cot A$ ?

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Answer 1 · 2015-10-04T15:40:18

$\frac{1}{sec A + 1} + \frac{1}{sec A - 1}$ $1 / (sec A + 1) + 1 / (Sec A - 1)$

Taking the Lowest common Multiple,

$\frac{sec A - 1 + sec A + 1}{sec A + 1} \cdot (sec A - 1)$ $(Sec A - 1 + Sec A + 1) / (Sec A +1) * (Sec A - 1)$

As you may be aware, $a^{2} - b^{2} = (a + b) \cdot (a - b)$ $a^2 - b^2 = (a + b) * (a - b)$

Simplifying, $\frac{2 sec A}{{sec}^{2} A - 1}$ $(2 Sec A) / (Sec^2 A - 1)$

Now ${sec}^{2} A - 1 = {tan}^{2} A = \frac{{sin}^{2} A}{{cos}^{2} A}$ $Sec^2 A - 1 = tan^2 A = Sin^2A / Cos^2A$
and $sec A = \frac{1}{cos A}$ $Sec A = 1 / Cos A$

Substituting,

$\frac{2}{cos A} \cdot \frac{{cos}^{2} A}{{sin}^{2} A} = 2 \cdot \frac{cos A}{{sin}^{2} A}$ $2 / Cos A * Cos^2A / Sin^2A = 2 * Cos A / Sin^2A$

which can be written as $2 \cdot \frac{cos A}{sin A} \cdot (\frac{1}{sin A})$ $2 * Cos A / Sin A * ( 1 / Sin A)$

Now $\frac{cos A}{sin A} = cot A and \frac{1}{sin A} = cos e c A$ $Cos A / Sin A = Cot A and 1 / Sin A = Cosec A$
Substituting, we get $2 cot A \cdot cos e c A$ $2 Cot A * Cosec A$

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How do I prove that $\frac{1}{sec A + 1} + \frac{1}{sec A - 1}$ $1/(sec A+1)+1/(sec A-1)$ = $2 csc A cot A$ $2 csc A cot A$ ?

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