Question

Period of the given trigonometric function $\tan x.\tan \left( {{{120}^0} - x} \right).\tan \left( {{{120}^0} + x} \right)$ is
$\left( a \right)\dfrac{\pi }{2} \\\ \left( b \right)\dfrac{\pi }{3} \\\ \left( c \right)\dfrac{{2\pi }}{3} \\\ \left( d \right)\pi \\\$

Answer 1

Hint-In this question, we use the concept of periodic function. Periodic function is a function that repeats its values in regular intervals or periods. A function is said to be periodic if there exists a positive real number T, $f\left( {x + T} \right) = f\left( x \right)$ and T is a period of function f(x).Period of such type of function, $ag\left( {bx + c} \right) + d$ is $\dfrac{T}{{\left| b \right|}}$ , where T is a fundamental period of g(x).

Complete step-by-step solution -
Let, $f\left( x \right) = \tan x.\tan \left( {{{120}^0} - x} \right).\tan \left( {{{120}^0} + x} \right)$
Now, we use $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{\sin \left( {{{120}^0} - x} \right).\sin \left( {{{120}^0} + x} \right)}}{{\cos \left( {{{120}^0} - x} \right).\cos \left( {{{120}^0} + x} \right)}}} \right)$
Multiply by 2 in numerator and denominator,
$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{2\sin \left( {{{120}^0} - x} \right).\sin \left( {{{120}^0} + x} \right)}}{{2\cos \left( {{{120}^0} - x} \right).\cos \left( {{{120}^0} + x} \right)}}} \right)$
Use trigonometric identity, $2\sin \left( {A + B} \right)\sin \left( {A - B} \right) = \cos \left( {2B} \right) - \cos \left( {2A} \right)$ and $2\cos \left( {A + B} \right)\cos \left( {A - B} \right) = \cos \left( {2A} \right) - \cos \left( {2B} \right)$.
$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{\cos \left( {2x} \right) - \cos \left( {{{240}^0}} \right)}}{{\cos \left( {{{240}^0}} \right) + \cos \left( {2x} \right)}}} \right)$
We know, $\cos \left( {{{240}^0}} \right) = \cos \left( {{{180}^0} + {{60}^0}} \right) = - \cos \left( {{{60}^0}} \right) = \dfrac{{ - 1}}{2}$

$$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{\cos \left( {2x} \right) + \dfrac{1}{2}}}{{ - \dfrac{1}{2} + \cos \left( {2x} \right)}}} \right) \\\ \Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{2\cos \left( {2x} \right) + 1}}{{2\cos \left( {2x} \right) - 1}}} \right) \\\$$

Now use trigonometric identity, $\cos \left( {2x} \right) = {\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)$ and ${\sin ^2}\left( x \right) + {\cos ^2}\left( x \right) = 1$

$$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{2\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + {{\sin }^2}x + {{\cos }^2}x}}{{2\left( {{{\cos }^2}x - {{\sin }^2}x} \right) - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}} \right) \\\ \Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{3{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x - 3{{\sin }^2}x}}} \right) \\\ \\\$$

Divide by ${\cos ^2}x$ in numerator and denominator,

$$\Rightarrow f\left( x \right) = \tan x.\left( {\dfrac{{3 - {{\tan }^2}x}}{{1 - 3{{\tan }^2}x}}} \right) \\\ \Rightarrow f\left( x \right) = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}} \\\$$

We know, $\tan \left( {3x} \right) = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$
$\Rightarrow f\left( x \right) = \tan \left( {3x} \right)$
We know tanx is a periodic function with fundamental period $\pi$ .
Now, we find period of $\tan \left( {3x} \right)$ so we use $ag\left( {bx + c} \right) + d$ is $\dfrac{T}{{\left| b \right|}}$ where T is a fundamental period of g(x).
Fundamental period of f(x), $T = \pi$ and value of b=3
Hence, period of f(x) is $\dfrac{\pi }{3}$
So, the correct option is (b).
Note-In such types of problems we use some important points to solve the question in an easy way like first we convert the question into simple forms like sine, cosine, tan etc. by using some trigonometric identities and then find the fundamental period of function.