# How do you verify the identify sinthetatantheta+costheta=sectheta?

Mar 9, 2018

See the proof below

#### Explanation:

We need

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\sec \theta = \frac{1}{\cos} \theta$

Therefore,

$L H S = \sin \theta \tan \theta + \cos \theta$

$= \sin \theta \cdot \sin \frac{\theta}{\cos} \theta + \cos \theta$

$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta}$

$= \frac{1}{\cos} \theta$

$= \sec \theta$

$= R H S$

$Q E D$

Mar 9, 2018

Apply the identities $\tan \left(\theta\right) = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$ along with the Pythagorean theorem.

#### Explanation:

Apply the identity $\tan \left(\theta\right) = \frac{\sin \theta}{\cos \theta}$:
$L . H . S . = \sin \theta \cdot \sin \frac{\theta}{\cos \theta} + \cos \theta$
$= {\sin}^{2} \frac{\theta}{\cos \theta} + \cos \theta$
$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta}$

Apply the Pythagorean theorem ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta}$
$= \frac{1}{\cos \theta}$

By the definition of secants $\frac{1}{\cos \theta} = \sec \theta$:
$= \sec \theta$
$= R . H . S$

Mar 9, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)tantheta=sintheta/costheta" and "sectheta=1/costheta"

•color(white)(x)sin^2theta+cos^2theta=1

$\text{Consider the left side}$

$\Rightarrow \sin \theta \tan \theta + \cos \theta$

$= \sin \theta \times \sin \frac{\theta}{\cos} \theta + \cos \theta$

$= {\sin}^{2} \frac{\theta}{\cos} \theta + {\cos}^{2} \frac{\theta}{\cos} \theta \leftarrow \text{common denominator } \cos \theta$

$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos} \theta$

$= \frac{1}{\cos} \theta = \sec \theta = \text{ right side "rArr" verified}$