Question

If \[\tan x = \dfrac{5}{{12}}\], and \[x\] lies in the third quadrant. Then what is the value of \[\cos\left( {\dfrac{x}{2}} \right)\]?
A. \[\dfrac{5}{{\sqrt {13} }}\]
B. \[\dfrac{5}{{\sqrt {26} }}\]
C. \[\dfrac{{ - 5}}{{\sqrt {13} }}\]
D. \[\sqrt {\dfrac{1}{{26}}} \]

Answer 1

Hint: First, calculate the range of \[\dfrac{x}{2}\]. Then use the trigonometric identity of \[\sec^{2}A\] and find the value of \[\cos x\] in the third quadrant. In the end, use the identity \[\cos 2A\] to reach the required answer.

Formula Used:
\[\sec^{2}A = \tan^{2}A + 1\]
\[\sec x = \dfrac{1}{{\cos x}}\]
\[\cos 2A = 2\cos^{2}A - 1\]

Complete step by step solution:
Given:
\[\tan x = \dfrac{5}{{12}}\], and \[x\] lies in the third quadrant.
Since \[x\] lies in the third quadrant.
So, \[180^{\circ} < x < 270^{\circ}\]
Divide each term by 2.
\[90^{\circ} < \dfrac{x}{2} < 135^{\circ}\]
Thus, \[\dfrac{x}{2}\] lies in the second quadrant.

Now apply the trigonometric identity \[\sec^{2}A = \tan^{2}A + 1\].
We get,
\[\sec^{2}x = \tan^{2}x + 1\]
Substitute the value of \[\tan x\] in the above equation.
\[\sec^{2}x = {\left( {\dfrac{5}{{12}}} \right)^2} + 1\]
\[ \Rightarrow \]\[\sec^{2}x = \dfrac{{25}}{{144}} + 1\]
\[ \Rightarrow \]\[\sec^{2}x = \dfrac{{169}}{{144}}\]
Take square root on both sides.
\[\sec x = \pm \dfrac{{13}}{{12}}\]
Since \[\sec x\] is negative in the third quadrant.
So, only possible value is \[\sec x = - \dfrac{{13}}{{12}}\].
Apply the trigonometric ratio \[\sec x = \dfrac{1}{{\cos x}}\].
\[\cos x = - \dfrac{{12}}{{13}}\]

Now apply the trigonometric identity \[\cos 2A = 2\cos^{2}A - 1\]
\[\cos x = 2\cos^{2}\dfrac{x}{2} - 1\]
\[ \Rightarrow \]\[2\cos^{2}\dfrac{x}{2} = 1 + \cos x\]
Substitute the value \[\cos x = - \dfrac{{12}}{{13}}\] in the above equation.
\[2\cos^{2}\dfrac{x}{2} = 1 + \left( {\dfrac{{ - 12}}{{13}}} \right)\]
\[ \Rightarrow \]\[2\cos^{2}\dfrac{x}{2} = \dfrac{1}{{13}}\]
Divide both sides by 2.
\[\cos^{2}\dfrac{x}{2} = \dfrac{1}{{26}}\]
Take square root on both sides.
\[\cos\dfrac{x}{2} = \sqrt {\dfrac{1}{{26}}} \]

Hence the correct option is B.

Note: Students often get confused with the values of trigonometric angles in each quadrant.
In the first quadrant, all trigonometric angles are positive.
In the second quadrant, the angles \[\sin x\] and \[\text{cosec} x\] are positive.
In the third quadrant, the angles \[\tan x\] and \[\cot x\] are positive.
In the fourth quadrant, the angles \[\cos x\] and \[\sec x\] are positive.