# How do I prove that 1/(sec A+1)+1/(sec A-1)=2 csc A cot A ?

Oct 4, 2015

$\frac{1}{\sec A + 1} + \frac{1}{S e c A - 1}$

Taking the Lowest common Multiple,

$\frac{S e c A - 1 + S e c A + 1}{S e c A + 1} \cdot \left(S e c A - 1\right)$

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

Simplifying, $\frac{2 S e c A}{S e {c}^{2} A - 1}$

Now $S e {c}^{2} A - 1 = {\tan}^{2} A = S {\in}^{2} \frac{A}{C} o {s}^{2} A$
and $S e c A = \frac{1}{C} o s A$

Substituting,

$\frac{2}{C} o s A \cdot C o {s}^{2} \frac{A}{S} {\in}^{2} A = 2 \cdot C o s \frac{A}{S} {\in}^{2} A$

which can be written as $2 \cdot C o s \frac{A}{S} \in A \cdot \left(\frac{1}{S} \in A\right)$

Now $C o s \frac{A}{S} \in A = C o t A \mathmr{and} \frac{1}{S} \in A = C o \sec A$
Substituting, we get $2 C o t A \cdot C o \sec A$