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

Question

James Adkins · Accepted Answer

#1 / (sec A + 1) + 1 / (Sec A - 1)# The Lowest Common Multiple is taken. #(Sec A - 1 + Sec A + 1) / (Sec A +1) * (Sec A - 1)# As you may be aware, #a^2 - b^2 = (a + b) * (a - b)# Simplifying, #(2 Sec A) / (Sec^2 A - 1)# Now #Sec^2 A - 1 = tan^2 A = Sin^2A / Cos^2A#
and #Sec A = 1 / Cos A# Changing, #2 / Cos A * Cos^2A / Sin^2A = 2 * Cos A / Sin^2A# which can be written as #2 * Cos A / Sin A * ( 1 / Sin A)# Now #Cos A / Sin A = Cot A and 1 / Sin A = Cosec A#
Substituting, we get #2 Cot A * Cosec A#

Zoe Austin · Answer

undefined To prove that $\frac{1}{(\sec A + 1)} + \frac{1}{(\sec A - 1)} = 2 \csc A \cot A$, we can start with the left side of the equation and manipulate it until we reach the expression on the right side.

Starting with the left side:

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

We'll first find a common denominator:

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

Combining the fractions:

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

Using the Pythagorean identity $\sec^2 A - 1 = \tan^2 A$:

$$ \frac{2\sec A}{\sec^2 A - 1} = \frac{2\sec A}{\tan^2 A} $$

Using the reciprocal identities $\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$:

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

Simplify:

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

$$ \frac{2}{\frac{\sin^2 A}{\cos A}} = 2 \csc A \cot A $$

Thus, we have proven that $\frac{1}{(\sec A + 1)} + \frac{1}{(\sec A - 1)} = 2 \csc A \cot A$.