What is the period of $f (θ) = tan (\frac{12 θ}{7}) - sec (\frac{7 θ}{6})$ $f(theta) = tan ( ( 12 theta)/7 )- sec ( ( 7 theta)/ 6 )$ ?

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What is the period of $f (θ) = tan (\frac{12 θ}{7}) - sec (\frac{7 θ}{6})$ $f(theta) = tan ( ( 12 theta)/7 )- sec ( ( 7 theta)/ 6 )$ ?

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Answer 1 · 2018-05-18T02:56:54

$84 π$ $84pi$

Explanation:

Period of $tan (\frac{12 t}{7}) - \to \frac{7 π}{12}$ $tan ((12t)/7) --> (7pi)/12$
Period of $sec (\frac{7 t}{6}) - \to 6 \frac{2 π}{7} = \frac{12 π}{7}$ $sec ((7t)/6) --> 6(2pi)/7 = (12pi)/7$
Period of f(t) --> least common multiple of $\frac{7 π}{12} and \frac{12 π}{7}$ $(7pi)/12 and (12pi)/7$

$\frac{7 π}{12}$ $(7pi)/12$ ......x...(12)(12) .... --> $84 π$ $84pi$
$\frac{12 π}{7}$ $(12pi)/7$ .......x......(7)(7) ..... --> $84 π$ $84pi$

Period of f(t) is $84 π$ $84pi$

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What is the period of $f (θ) = tan (\frac{12 θ}{7}) - sec (\frac{7 θ}{6})$ $f(theta) = tan ( ( 12 theta)/7 )- sec ( ( 7 theta)/ 6 )$ ?

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Explanation:

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What is the period of f(θ)=tan(12θ7)−sec(7θ6)f(theta) = tan ( ( 12 theta)/7 )- sec ( ( 7 theta)/ 6 ) ?

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Explanation:

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What is the period of $f (θ) = tan (\frac{12 θ}{7}) - sec (\frac{7 θ}{6})$ $f(theta) = tan ( ( 12 theta)/7 )- sec ( ( 7 theta)/ 6 )$ ?