# How do you solve: tanθ + secθ = 1 ?

May 23, 2018

tanθ + secθ = 1....

Now
We know ${\sec}^{2} \theta - {\tan}^{2} \theta = 1$
$\implies \sec \theta - \tan \theta = \frac{1}{\sec \theta + \tan \theta} = \frac{1}{1}$

$\implies \sec \theta - \tan \theta = 1. \ldots . \left[2\right]$

Adding  and  we get

$2 \sec \theta = 2$

$\implies \cos \theta = 1$

$\implies \theta = 2 n \pi \text{ where "n inZZ}$

May 23, 2018

$\theta = \pm k \pi , \frac{3 \pi}{2} + 2 k \pi , - \frac{\pi}{2} - 2 k \pi$

#### Explanation:

.

$\tan \theta + \sec \theta = 1$

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

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

$\sin \theta + 1 = \cos \theta$

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

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

$2 {\sin}^{2} \theta + 2 \sin \theta = 0$

$2 \left({\sin}^{2} \theta + \sin \theta\right) = 0$

${\sin}^{2} \theta + \sin \theta = 0$

$\sin \theta \left(\sin \theta + 1\right) = 0$

$\sin \theta = 0 , \therefore \theta = \pm k \pi$

$\sin \theta + 1 = 0 , \therefore \sin \theta = - 1 , \therefore \theta = \frac{3 \pi}{2} + 2 k \pi , - \frac{\pi}{2} - 2 k \pi$

May 23, 2018

$\frac{\pi}{2} \pm 2 \pi n , \pi \pm 2 \pi n$; $n \in \mathbb{Z}$

#### Explanation:

We have: $\tan \left(\theta\right) + \sec \left(\theta\right) = 1$

$R i g h t a r r o w \frac{\sin \left(\theta\right)}{\cos \left(\theta\right)} + \frac{1}{\cos \left(\theta\right)} = 1$

$R i g h t a r r o w \frac{\sin \left(\theta\right) + 1}{\cos \left(\theta\right)} = 1$

$R i g h t a r r o w \sin \left(\theta\right) + 1 = \cos \left(\theta\right)$

$R i g h t a r r o w \sin \left(\theta\right) - \cos \left(\theta\right) = - 1$

Then, let's consider $\sin \left(\theta\right) - \cos \left(\theta\right) = R \cos \left(\theta + \alpha\right)$.

We can expand the right-hand side using the compound angle identity for $\cos \left(\theta\right)$:

$R i g h t a r r o w \sin \left(\theta\right) - \cos \left(\theta\right) = R \left(\cos \left(\theta\right) \cos \left(\alpha\right) - \sin \left(\theta\right) \sin \left(\alpha\right)\right)$

$R i g h t a r r o w \sin \left(\theta\right) - \cos \left(\theta\right) = R \cos \left(\theta\right) \cos \left(\alpha\right) - R \sin \left(\theta\right) \sin \left(\alpha\right)$

$R i g h t a r r o w \sin \left(\theta\right) - \cos \left(\theta\right) = R \cos \left(\alpha\right) \cos \left(\theta\right) - R \sin \left(\alpha\right) \sin \left(\theta\right)$

If we compare the coefficients of $\sin \left(\theta\right)$ and $\cos \left(\theta\right)$, we get:

$R i g h t a r r o w 1 = - R \sin \left(\alpha\right) R i g h t a r r o w R \sin \left(\alpha\right) = - 1 \text{ " " " " }$ $\left(i\right)$

and

$R i g h t a r r o w - 1 = R \cos \left(\alpha\right) R i g h t a r r o w R \cos \left(\alpha\right) = - 1 \text{ " " " " }$ $\left(i i\right)$

If we divide $\left(i\right)$ by $\left(i i\right)$, we get:

$R i g h t a r r o w \frac{R \sin \left(\alpha\right)}{R \cos \left(\alpha\right)} = \frac{- 1}{- 1}$

$R i g h t a r r o w \tan \left(\alpha\right) = 1$

$R i g h t a r r o w \alpha = \frac{\pi}{4}$

If we sum the squares of $\left(i\right)$ and $\left(i i\right)$, we get:

$R i g h t a r r o w {R}^{2} {\sin}^{2} \left(\alpha\right) + {R}^{2} {\cos}^{2} \left(\alpha\right) = 1 + 1$

$R i g h t a r r o w {R}^{2} \left({\cos}^{2} \left(\alpha\right) + {\sin}^{2} \left(\alpha\right)\right) = 2$

$R i g h t a r r o w {R}^{2} = 2$

$R i g h t a r r o w R = \sqrt{2}$

So, we can write $\sin \left(\theta\right) - \cos \left(\theta\right)$ in the form $\sqrt{2} \cos \left(\theta + \frac{\pi}{4}\right)$, i.e. in our original problem we get $\sqrt{2} \cos \left(\theta + \frac{\pi}{4}\right) = - 1$:

$R i g h t a r r o w \cos \left(\theta + \frac{\pi}{4}\right) = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$

Now, let the reference angle be $\cos \left(\theta\right) = \frac{\sqrt{2}}{2} R i g h t a r r o w \theta = \frac{\pi}{4}$.

But the value of $\cos \left(\theta + \frac{\pi}{4}\right)$ is negative, so $\theta$ must be located in either the second or third quadrant:

$R i g h t a r r o w \theta + \frac{\pi}{4} = \pi - \frac{\pi}{4} , \pi + \frac{\pi}{4}$

$R i g h t a r r o w \theta + \frac{\pi}{4} = \frac{3 \pi}{4} , \frac{5 \pi}{4}$

$R i g h t a r r o w \theta = \frac{\pi}{2} , \pi$

As you have not specified a domain that $\theta$ must be located within, we can give the general solutions to the equation as:

$\therefore \theta = \frac{\pi}{2} \pm 2 \pi n , \pi \pm 2 \pi n$; $n \in \mathbb{Z}$

May 23, 2018

$t = k \pi$
$t = \frac{3 \pi}{2} + 2 k \pi$

#### Explanation:

$\sin \frac{t}{\cos t} + \frac{1}{\cos} t = 1$
$\frac{\sin t + 1}{\cos t} = 1$
$\sin t + 1 = \cos t$
$\sin t - \cos t = - 1$
Use trig identity: $\sin t - \cos t = - \sqrt{2} \cos \left(t + \frac{\pi}{4}\right)$
In this case:
$\sin t - \cos t = - \sqrt{2} \cos \left(t + \frac{\pi}{4}\right) = - 1$
$\cos \left(t + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Trig table and unit circle give 2 solutions for t:
$\left(t + \frac{\pi}{4}\right) = \pm \frac{\pi}{4}$
a. $t + \frac{\pi}{4} = \frac{\pi}{4}$
t = 0, $t = \pi$, and $t = 2 \pi$
General answer --> $t = k \pi$
b. t + pi/4 = - pi/4
$t = - \frac{\pi}{2} + 2 k \pi$, or $t = \frac{3 \pi}{2} + 2 k \pi$ (co-terminal)