Question #1a94c

May 14, 2017

See below.

Explanation:

Prove: ${\tan}^{2} \frac{x}{{\tan}^{2} x + 1} = {\sin}^{2} x$

Let us start on the left-hand side (LHS) and work toward the right-hand side (RHS).

Consider Pythagorean's identity:
${\sin}^{2} x + {\cos}^{2} x = 1$

which we can divide by ${\cos}^{2} x$ to get:
${\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} x = \frac{1}{\cos} ^ 2 x$

Note that $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$:
${\tan}^{2} x + 1 = {\sec}^{2} x$, which is most applicable to this problem

We can substitute ${\tan}^{2} x + 1 = {\sec}^{2} x$ into the LHS:
$L H S = {\tan}^{2} \frac{x}{\sec} ^ 2 x$

Note that $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$:
$L H S = {\tan}^{2} \frac{x}{\frac{1}{\cos} ^ 2 x}$

$= \frac{{\sin}^{2} \frac{x}{\cos} ^ 2 x}{\frac{1}{\cos} ^ 2 x}$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x \cdot {\cos}^{2} x$

$= {\sin}^{2} x = R H S$

Therefore, ${\tan}^{2} \frac{x}{{\tan}^{2} x + 1} = {\sin}^{2} x$