Question

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

Answer 1

`1 / (sec A + 1) + 1 / (Sec A - 1)`

Taking the Lowest common Multiple,

`(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`

Substituting,

`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`