The period of the function $tan (3 x + 5)$ is

\frac{2 π}{3}

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\frac{π}{6}

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\frac{π}{3}

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none of these

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Solution

The correct option is D $\frac{π}{3}$
Let, $f (x) = tan (3 x + 5)$
By property of Periodic function.
period of $f (x) =$ period of $f (x + λ), λ ϵ R$
$∴$ period of $tan (3 x) =$ period of $tan (3 x + λ), λ ϵ R$
$∴$ period of $tan (3 x + 5) = \frac{π}{3}$
[Period of $tan x = π, ∴$ period of $tan λ x = \frac{π}{| λ |} a n d p e r i o d o f tan (x) = p e r i o d o f tan (x + a)$ ]

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