Question

How do you prove `[(1)/(1-sinx)]+[(1)/(1+sinx)]=2sec^2x`?

Answer 1

The formula can be proven by applying: 1) Least common multiple; 2) applying the trigonometric entity `sin^2x + cos^2x=1`

Explanation:

Head

Key-relation : `sin^2x + cos^2x=1`

Key-concept: Least common multiple; when no common multiples, just multiply the terms in the denominator.

Calculation

The above formula can be proven by transforming left side to right side:

`1/(1-sin x)+1/(1+sin x)= (1+sin x + 1-sin x)/((1+sinx)(1-sinx))`

To arrive to right-hand side, just divide the denominator to `(1+sinx)(1-sinx)`, the least common multiple, and multiply the numerator to the remaining, since they are all 1, just put the value.

By simple algebra and make use of `(a-b)(a+b)=a^2 - b^2`, it can be seen from normal multiplication.

`(1+sin x + 1-sin x)/((1+sinx)(1-sinx))= 2/(1-sin^2x)`

Finally apply: `sin^2x + cos^2x=1`, which gives out `cos^2x=1 - sin^2x`

`2/(1-sin^2x)=2/cos^2x=2*(1/cosx)^2`

To finish, remember that `secx=1/cosx`, hence:

`2*(1/cosx)^2=2sec^2x`