Problem of the week #62 - June 3rd, 2013

[Jameson]

Demonstrate the following identity:

$\displaystyle \frac {\tan x} {\sec x + 1} = \frac {\sec x - 1} {\tan x}$
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[Jameson]

Congratulations to the following members for their correct solutions:

1) anemone
2) Prove It
3) soroban
4) MarkFL
5) Sudharaka

Solution (from Prove It):

Spoiler

\(\displaystyle \displaystyle \begin{align*} \frac{\tan{(x)}}{\sec{(x)} + 1} &= \frac{\tan{(x)}\left[ \sec{(x)} - 1 \right]}{\left[ \sec{(x)} + 1 \right] \left[ \sec{(x)} - 1 \right] } \\ &= \frac{\tan{(x)}\left[ \sec{(x)} - 1 \right] }{ \sec^2{(x)} - 1 } \\ &= \frac{\tan{(x)}\left[ \sec{(x)} - 1 \right] }{ \tan^2{(x)}} \\ &= \frac{\sec{(x)} - 1}{\tan{(x)}} \end{align*}\)

Alternate solution (from MarkFL):

Spoiler

Beginning with the left side of the identity, and utilizing double-angle identities for sine and cosine, we may write:

\(\displaystyle \frac{\tan(x)}{\sec(x)+1}\cdot\frac{\cos(x)}{\cos(x)}=\frac{\sin(x)}{1+\cos(x)}=\frac{\sin\left(2 \cdot\frac{x}{2} \right)}{1+\cos\left(2\cdot\frac{x}{2} \right)}=\)

\(\displaystyle \frac{2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)}{2\cos^2\left(\frac{x}{2} \right)}=\frac{2\sin\left(\frac{x}{2} \right)}{2\cos\left(\frac{x}{2} \right)}\cdot\frac{\sin\left(\frac{x}{2} \right)}{\sin\left(\frac{x}{2} \right)}=\)

\(\displaystyle \frac{2\sin^2\left(\frac{x}{2} \right)}{2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)}=\frac{1-\cos(x)}{\sin(x)}\cdot\frac{\sec(x)}{\sec(x)}= \frac{\sec(x)-1}{\tan(x)}\)