# What is the period of f(theta) = tan ( ( 15 theta)/7 )- sec ( ( 5 theta)/ 6 ) ?

##### 1 Answer

Period $P = \frac{84 \pi}{5} = 52.77875658$

#### Explanation:

The given $f \left(\theta\right) = \tan \left(\frac{15 \theta}{7}\right) - \sec \left(\frac{5 \theta}{6}\right)$

For $\tan \left(\frac{15 \theta}{7}\right)$, period ${P}_{t} = \frac{\pi}{\frac{15}{7}} = \frac{7 \pi}{15}$

For $\sec \left(\frac{5 \theta}{6}\right)$, period ${P}_{s} = \frac{2 \pi}{\frac{5}{6}} = \frac{12 \pi}{5}$

To get the period of $f \left(\theta\right) = \tan \left(\frac{15 \theta}{7}\right) - \sec \left(\frac{5 \theta}{6}\right)$,
We need to obtain the LCM of the ${P}_{t}$ and ${P}_{s}$

The solution

Let $P$ be the required period
Let $k$ be an integer such that $P = k \cdot {P}_{t}$
Let $m$ be an integer such that $P = m \cdot {P}_{s}$

$P = P$
$k \cdot {P}_{t} = m \cdot {P}_{s}$
$k \cdot \frac{7 \pi}{15} = m \cdot \frac{12 \pi}{5}$

Solving for $\frac{k}{m}$

$\frac{k}{m} = \frac{15 \left(12\right) \pi}{5 \left(7\right) \pi}$

$\frac{k}{m} = \frac{36}{7}$

We use $k = 36$ and $m = 7$
so that
$P = k \cdot {P}_{t} = 36 \cdot \frac{7 \pi}{15} = \frac{84 \pi}{5}$

also

$P = m \cdot {P}_{s} = 7 \cdot \frac{12 \pi}{5} = \frac{84 \pi}{5}$

Period $P = \frac{84 \pi}{5} = 52.77875658$

Kindly see the graph and observe two points to verify for the period

God bless....I hope the explanation is useful