Trig Identities #2

(1 + sin x)/(cos x) + (cos x)/(1 + sin x) = (4 sin x)/(sin 2x) {nl} Show that left equals right, and sin 2x is a double-angle identity. I despise Trig, haha.

\(\begin{array}{rcll} \frac{ 1 + \sin{( x )} } { \cos {( x )} } + \frac{ \cos {( x )} } { 1 + \sin {( x )} } &\overset{?}{=}& \frac{ 4 \sin {( x )} } { \sin {( 2x )} } \qquad & | \qquad \sin {( 2x )} = 2\cdot \sin {( x )}\cos {( x )} \\\\ &\overset{?}{=}& \frac{ 4 \sin {( x )} } { 2\cdot \sin {( x )}\cos {( x )} } \\\\ &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 1 + \sin{( x )} } { \cos {( x )} } + \frac{ \cos {( x )} } { 1 + \sin {( x )} } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ [1 + \sin{( x )}]^2 + [\cos^2 {( x )}] } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 1+2\sin{( x )} + \sin{( x )}^2 + \cos^2 {( x )} } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \qquad & | \qquad \sin{( x )}^2 + \cos^2 {( x )} = 1 \\\\ \frac{ 1+2\sin{( x )} + 1} { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2+2\sin{( x )} } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2\cdot[ 1+\sin{( x )} ] } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2 } { \cos {( x )} } &=& \frac{ 2 } { \cos {( x )} } \end{array}\)