# Question #4352a

Mar 26, 2017

$1$

#### Explanation:

If $3 {\sin}^{2} x - {\sin}^{4} x = 1$ then ${\sin}^{2} x = \frac{3 - \sqrt{5}}{2}$

and

${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - \frac{3 - \sqrt{5}}{2}$

then

${\tan}^{2} x + {\tan}^{4} x = \frac{\frac{3 - \sqrt{5}}{2}}{1 - \frac{3 - \sqrt{5}}{2}} + {\left(\frac{\frac{3 - \sqrt{5}}{2}}{1 - \frac{3 - \sqrt{5}}{2}}\right)}^{2} = 1$

Mar 27, 2017

$3 - {\sin}^{2} x$

#### Explanation:

Develop the left side of the equation:
$3 {\sin}^{2} x - {\sin}^{4} x = 1$
${\sin}^{2} x \left(3 - {\sin}^{2} x\right) = 1$
From this equation we get:
$\frac{1}{\sin} ^ 2 x = 3 - {\sin}^{2} x$ (1)

Develop the right side:
${\tan}^{2} x + {\tan}^{4} x = {\tan}^{2} x \left(1 + {\tan}^{2} x\right) = {\tan}^{2} x \left(\frac{1}{\cos} ^ 2 x\right) =$
$= \frac{{\cos}^{2} x}{{\sin}^{2} x} \left(\frac{1}{{\cos}^{2} x}\right) = \frac{1}{{\sin}^{2} x} \left(2\right)$
Compare (1) and (2) -->
${\tan}^{2} x + {\tan}^{4} x = 3 - {\sin}^{2} x$

Mar 27, 2017

$3 {\sin}^{2} x - {\sin}^{4} x = 1$

Dividing bothsides by ${\cos}^{4} x$

$\implies \frac{3 {\sin}^{2} x}{\cos} ^ 4 x - {\sin}^{4} \frac{x}{\cos} ^ 4 x = \frac{1}{\cos} ^ 4 x$

$\implies 3 {\tan}^{2} x {\sec}^{2} x - {\tan}^{4} x = {\sec}^{4} x$

$\implies 3 {\tan}^{2} x \left(1 + {\tan}^{2} x\right) - {\tan}^{4} x = {\left(1 + {\tan}^{2} x\right)}^{2}$

$\implies 3 {\tan}^{2} x + 3 {\tan}^{4} x - {\tan}^{4} x = 1 + 2 {\tan}^{2} x + {\tan}^{4} x$

$\implies 3 {\tan}^{2} x - 2 {\tan}^{2} x + 3 {\tan}^{4} x - {\tan}^{4} x - {\tan}^{4} x = 1$

$\implies {\tan}^{2} x + {\tan}^{4} x = 1$