# How do you verify (sin^3 t + cos^3 t)/(cos2t) = (sec^2 t - tant)/(sect - tant*sect)?

Nov 26, 2015

Apply identities, and use the difference of square and sum of cubes formulas.

#### Explanation:

We will be using the following:
$\cos \left(2 t\right) = {\cos}^{2} \left(t\right) - {\sin}^{2} \left(t\right)$
${\sin}^{2} \left(t\right) + {\cos}^{2} \left(t\right) = 1$
$\sec \left(t\right) = \frac{1}{\cos} \left(t\right)$
$\tan \left(t\right) = \sin \frac{t}{\cos} \left(t\right)$
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Starting from the right hand side:

$\frac{{\sec}^{2} \left(t\right) - \tan \left(t\right)}{\sec \left(t\right) - \tan \left(t\right) \sec \left(t\right)} = \frac{\frac{1}{\cos} ^ 2 \left(t\right) - \sin \frac{t}{\cos} \left(t\right)}{\frac{1}{\cos} \left(t\right) - \sin \frac{t}{\cos} ^ 2 \left(t\right)}$

$= \frac{\frac{1}{\cos} ^ 2 \left(t\right) - \sin \frac{t}{\cos} \left(t\right)}{\frac{1}{\cos} \left(t\right) - \sin \frac{t}{\cos} ^ 2 \left(t\right)} \cdot {\cos}^{2} \frac{t}{\cos} ^ 2 \left(t\right)$

$= \frac{1 - \sin \left(t\right) \cos \left(t\right)}{\cos \left(t\right) - \sin \left(t\right)}$

$= \frac{{\sin}^{2} \left(t\right) + {\cos}^{2} \left(t\right) - \sin \left(t\right) \cos \left(t\right)}{\cos \left(t\right) - \sin \left(t\right)}$

$= \frac{{\sin}^{2} \left(t\right) + {\cos}^{2} \left(t\right) - \sin \left(t\right) \cos \left(t\right)}{\cos \left(t\right) - \sin \left(t\right)} \cdot \frac{\sin \left(t\right) + \cos \left(t\right)}{\sin \left(t\right) + \cos \left(t\right)}$

= ((cos(t)+sin(t))(sin^2(t) - sin(t)cos(t) + cos^2(t)))/((cos(t)+sin(t))(cos(t)-sin(t))

$= \frac{{\sin}^{3} \left(t\right) + {\cos}^{3} \left(t\right)}{{\cos}^{2} \left(t\right) - {\sin}^{2} \left(t\right)}$

$= \frac{{\sin}^{3} \left(t\right) + {\cos}^{3} \left(t\right)}{\cos} \left(2 t\right)$