The given f(theta)=tan((15theta)/7)-sec ((5theta)/6)f(θ)=tan(15θ7)−sec(5θ6)
For tan ((15theta)/7)tan(15θ7), period P_t=pi/(15/7)=(7pi)/15Pt=π157=7π15
For sec ((5theta)/6)sec(5θ6), period P_s=(2pi)/(5/6)=(12pi)/5Ps=2π56=12π5
To get the period of f(theta)=tan((15theta)/7)-sec ((5theta)/6)f(θ)=tan(15θ7)−sec(5θ6),
We need to obtain the LCM of the P_tPt and P_sPs
The solution
Let PP be the required period
Let kk be an integer such that P=k*P_tP=k⋅Pt
Let mm be an integer such that P=m*P_sP=m⋅Ps
P=PP=P
k*P_t=m*P_sk⋅Pt=m⋅Ps
k*(7pi)/15=m*(12pi)/5k⋅7π15=m⋅12π5
Solving for k/mkm
k/m=(15(12)pi)/(5(7)pi)km=15(12)π5(7)π
k/m=36/7km=367
We use k=36k=36 and m=7m=7
so that
P=k*P_t=36*(7pi)/15=(84pi)/5P=k⋅Pt=36⋅7π15=84π5
also
P=m*P_s=7*(12pi)/5=(84pi)/5P=m⋅Ps=7⋅12π5=84π5
Period P=(84pi)/5=52.77875658P=84π5=52.77875658
Kindly see the graph and observe two points to verify for the period
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God bless....I hope the explanation is useful
