Question

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

Answer 1

Period `P=(84pi)/5=52.77875658`

Explanation:

The given `f(theta)=tan((15theta)/7)-sec ((5theta)/6)`

For `tan ((15theta)/7)`, period `P_t=pi/(15/7)=(7pi)/15`

For `sec ((5theta)/6)`, period `P_s=(2pi)/(5/6)=(12pi)/5`

To get the period of `f(theta)=tan((15theta)/7)-sec ((5theta)/6)`,
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*P_t`
Let `m` be an integer such that `P=m*P_s`

`P=P`
`k*P_t=m*P_s`
`k*(7pi)/15=m*(12pi)/5`

Solving for `k/m`

`k/m=(15(12)pi)/(5(7)pi)`

`k/m=36/7`

We use `k=36` and `m=7`
so that
`P=k*P_t=36*(7pi)/15=(84pi)/5`

also

`P=m*P_s=7*(12pi)/5=(84pi)/5`

Period `P=(84pi)/5=52.77875658`

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

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