We know that the Principal Period of #tan# & #cos# are, #pi# & #2pi#, resp., which means that,
#AAx in RR, tan (x+pi)=tanx...........(1),# and, #cos(x+2pi)=cosx.............(2)#
We will use this fact in solving this problem.
Put #g(theta)=tan(13theta/4)# and #h(theta)=cos(2theta/5)# to give #f(theta)=g(theta)-h(theta)#
By #(1), tan(13theta/4)=tan(13theta/4+pi)=tan{13/4(theta+4pi/13)},#, i.e.,
#g(theta)=g(theta+4pi/13),# hence, the period #p_1# of fun. #g# is #p_1=4pi/13#
On the same line, we can get the period #p_2# of fun. #h, p_2=2pi/(2/5)=5pi.#
If we can find #l,m in NN# such that #l*p_1=m*p_2=p......(3)#,then #f=g-h# will be periodic and #p# will be its Prin. Period.
Now, #l*p_1=m*p_2 rArr l(4pi/13)=m(5pi) rArr 4l=65m.# Therefore, taking #l=65, and, m=4#, we see that #l,m in NN, and, (3)# is satisfied, bcz., #65*p_1=4*p_2,=65*4pi/13=4*5pi=20pi=p#
Therefore, #f=g-h# is periodic with Prin. Prd. #p=20pi.#
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