Question

If $2\sin \left( {x + \dfrac{\pi }{3}} \right) = \cos \left( {x - \dfrac{\pi }{6}} \right)$, then $\tan x =$
A.$- \sqrt 3$
B.$\sqrt 3$
C.$- 1/\sqrt 3$
D.$1/\sqrt 3$

Answer 1

Here, we will expand the trigonometric functions using the basic identities of the trigonometric functions. We will simplify both sides of the equation and convert it in the form of $\tan x.$. On further simplification, we will get the required value of $\tan x$.The tangent of an angle is equal to the ratio of sine of that angle to the cosine of the same angle.

Formula Used: We will use the following formulas/identities:
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$

Complete step-by-step answer:
We will expand sine function and cosine function using the trigonometric identities, $\sin \left( {A + B} \right)$ and $\cos \left( {A - B} \right)$. Therefore, we get
$\Rightarrow 2\left( {\sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}} \right) = \cos x\cos \dfrac{\pi }{6} + \sin x\sin \dfrac{\pi }{6}$
We know that the value of $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$, $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$, $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$. Therefore, we get
$\Rightarrow 2\left( {\sin x \times \dfrac{1}{2} + \cos x \times \dfrac{{\sqrt 3 }}{2}} \right) = \cos x \times \dfrac{{\sqrt 3 }}{2} + \sin x \times \dfrac{1}{2}$
As all the terms have $\dfrac{1}{2}$ common.
So, canceling out $\dfrac{1}{2}$, we get
$\Rightarrow 2\left( {\sin x + \cos x \times \sqrt 3 } \right) = \cos x \times \sqrt 3 + \sin x$
$\Rightarrow 2\sin x + 2\cos x \times \sqrt 3 = \cos x \times \sqrt 3 + \sin x$
Adding and subtracting the like terms, we get
$\Rightarrow 2\sin x - \sin x = \cos x \times \sqrt 3 - 2\cos x \times \sqrt 3$
$\Rightarrow \sin x = - \cos x \times \sqrt 3$
Dividing both sides by $\cos x$, we get
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} = - \sqrt 3$
We know that the tangent function is the ratio of the sine function to the cosine function.
Therefore, we get
$\Rightarrow \tan x = - \sqrt 3$
Hence, $\tan x$ is equal to $- \sqrt 3$.
So, option A is the correct option.

Note: We should know the different properties of the trigonometric function and with the help of this concept this question can be easily solved. Trigonometry is the study of the relationship between the angles and sides of a right triangle. Right Triangle is a triangle where one of its interior angles is a right angle (90 degrees). The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called base.
$\begin{array}{l}\sin {\rm{ }}A{\rm{ }} = {\rm{ }}\dfrac{{side\,{\rm{ }}\,opposite\,{\rm{ }}to\,{\rm{ }}angle\,{\rm{ }}A}}{{hypotenuse}}\\\\\cos {\rm{ }}A{\rm{ }} = \dfrac{{side\,{\rm{ }}adjacent\,{\rm{ }}to\,{\rm{ }}angle\,{\rm{ }}A}}{{hypotenuse}}\\\\\tan {\rm{ }}A{\rm{ }} = {\rm{ }}\dfrac{{side\,{\rm{ }}opposite\,{\rm{ }}to\,{\rm{ }}angle\,{\rm{ }}A}}{{side\,{\rm{ }}adjacent\,{\rm{ }}to\,{\rm{ }}angle\,{\rm{ }}A}} = \dfrac{{\sin A}}{{\cos A}}\end{array}$
$\cot {\rm{ }}A = \dfrac{1}{{\tan {\rm{ }}A}},\sec {\rm{ }}A = \dfrac{1}{{\cos {\rm{ }}A}},\cos ec{\rm{ }}A = \dfrac{1}{{\sin {\rm{ }}A}}$