Question

How do you prove: `cos (x)/(1+sin (x)) + (1+sin (x))/cos (x) = 2 sec (x)`?

Answer 1

Let's take the first member:

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

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

`=(2(1+sinx))/((1+sinx)cosx)=2/cosx=2secx`.

(Remember that: `sin^2x+cos^2x=1`).

Answer 2

First of all, the equality is possible only if `1+sin(x)!=0` and `cos(x)!=0`.
The corresponding conditions on an angle `x`, that we assume prior to proving the equality, are:
`x!=pi/2+pi*n`

Let's try to simplify it in straight forward fashion.

First, bring the left part to a common denominator:
`{cos^2(x)+[1+sin(x)]^2}/{[1+sin(x)]*cos(x)}`

Transform the numerator, leave the denominator:
`{cos^2(x)+1+2sin(x)+sin^2(x)}/{[1+sin(x)]*cos(x)}`

Recall that `sin^2(x)+cos^2(x)=1` for any `x`.
Therefore, our expression looks like
`{2+2sin(x)}/{[1+sin(x)]*cos(x)}={2[1+sin(x)]}/{[1+sin(x)]*cos(x)}`

We can safely reduce the last fraction by `1+sin(x)` (recall, we assumed that `x!=pi/2+pi*n` and, consequently, `1+sin(x)!=0`) getting, using the definition of a function `sec(x)`,
`2/cos(x)=2sec(x)`