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

Apr 15, 2015

Let's take the first member:

$\cos \frac{x}{1 + \sin \left(x\right)} + \frac{1 + \sin \left(x\right)}{\cos} \left(x\right) = \frac{{\cos}^{2} x + {\left(1 + \sin x\right)}^{2}}{\left(1 + \sin x\right) \cos x} =$

$= \frac{{\cos}^{2} x + 1 + 2 \sin x + {\sin}^{2} x}{\left(1 + \sin x\right) \cos x} = \frac{2 + 2 \sin x}{\left(1 + \sin x\right) \cos x} =$

$= \frac{2 \left(1 + \sin x\right)}{\left(1 + \sin x\right) \cos x} = \frac{2}{\cos} x = 2 \sec x$.

(Remember that: ${\sin}^{2} x + {\cos}^{2} x = 1$).

Apr 15, 2015

First of all, the equality is possible only if $1 + \sin \left(x\right) \ne 0$ and $\cos \left(x\right) \ne 0$.
The corresponding conditions on an angle $x$, that we assume prior to proving the equality, are:
$x \ne \frac{\pi}{2} + \pi \cdot n$

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

First, bring the left part to a common denominator:
$\frac{{\cos}^{2} \left(x\right) + {\left[1 + \sin \left(x\right)\right]}^{2}}{\left[1 + \sin \left(x\right)\right] \cdot \cos \left(x\right)}$

Transform the numerator, leave the denominator:
$\frac{{\cos}^{2} \left(x\right) + 1 + 2 \sin \left(x\right) + {\sin}^{2} \left(x\right)}{\left[1 + \sin \left(x\right)\right] \cdot \cos \left(x\right)}$

Recall that ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ for any $x$.
Therefore, our expression looks like
$\frac{2 + 2 \sin \left(x\right)}{\left[1 + \sin \left(x\right)\right] \cdot \cos \left(x\right)} = \frac{2 \left[1 + \sin \left(x\right)\right]}{\left[1 + \sin \left(x\right)\right] \cdot \cos \left(x\right)}$

We can safely reduce the last fraction by $1 + \sin \left(x\right)$ (recall, we assumed that $x \ne \frac{\pi}{2} + \pi \cdot n$ and, consequently, $1 + \sin \left(x\right) \ne 0$) getting, using the definition of a function $\sec \left(x\right)$,
$\frac{2}{\cos} \left(x\right) = 2 \sec \left(x\right)$