Question #58cc8

Jul 27, 2016

see explanation

Explanation:

Rewrite as:

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

Divide both sides by ${\cos}^{2} \left(x\right)$:

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

Take ${\sin}^{2} x + {\cos}^{2} x = 1$ and divide by ${\cos}^{2} x$ to get:

${\tan}^{2} x + 1 = {\sec}^{2} x$

$\implies 1 + {\cos}^{2} \left(x\right) = 1 + {\tan}^{2} \left(x\right)$

$\implies {\cos}^{2} \left(x\right) = {\tan}^{2} \left(x\right)$

Divide by ${\cos}^{2} \left(x\right)$ both sides:
$1 = {\tan}^{2} \left(x\right) \cdot \frac{1}{\cos} ^ 2 \left(x\right)$
or
$1 = {\tan}^{2} \left(x\right) \cdot {\sec}^{2} \left(x\right)$

$\therefore$ statement becomes

${\tan}^{2} \left(x\right) \left(1 + {\tan}^{2} \left(x\right)\right) = 1$

$\implies {\tan}^{2} \left(x\right) + {\tan}^{4} \left(x\right) = 1$ as required