Question

If we have the value of x as $0 < x < \pi $, and $\cos x + \sin x = \dfrac{1}{2}$, then find the value of $\tan x$?
A). $\dfrac{{\left( { - 4 \pm \sqrt 7 } \right)}}{3}$
B). $\dfrac{{\left( {1 + \sqrt 7 } \right)}}{4}$
C). $\dfrac{{\left( {1 - \sqrt 7 } \right)}}{4}$
$\dfrac{{\left( { - 1 \pm \sqrt 7 } \right)}}{4}$

Answer 1

Hint: : In the given problem we need to find value of $\tan x$ so we will try to convert $\cos x + \sin x = \dfrac{1}{2}$ in terms of tangent using trigonometric identities. We will use the substitution method to solve the given question and from that, we will obtain a quadratic equation and from that, we will find the value of $\tan x$.
Formulae used:
$\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
$\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
The above written identities are trigonometric ratios of angle $2x$, so will convert them to angle $x$.

Complete step-by-step solution:
Given, $\cos x + \sin x = \dfrac{1}{2}$
Since, we know that $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$, therefore $\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$
And, $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$, therefore $\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$
Let us substitute the value of $\cos x$ and $\sin x$ in $\cos x + \sin x = \dfrac{1}{2}$
$\therefore \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} + \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} = \dfrac{1}{2}$
Let $\tan \dfrac{x}{2} = t$
$ \Rightarrow \dfrac{{1 - {t^2}}}{{1 + {t^2}}} + \dfrac{{2t}}{{1 + {t^2}}} = \dfrac{1}{2}$
Take L.C.M
\[ \Rightarrow \dfrac{{1 - {t^2} + 2t}}{{1 + {t^2}}} = \dfrac{1}{2}\]
On cross-multiplication, we get
\[ \Rightarrow 2\left( {1 - {t^2} + 2t} \right) = 1 + {t^2}\]
$ \Rightarrow 2 - 2{t^2} + 4t = 1 + {t^2}$
Shift all terms on one side
$ \Rightarrow 2{t^2} + {t^2} - 4t - 2 + 1 = 0$
$ \Rightarrow 3{t^2} - 4t - 1 = 0$
The above written equation does not exists real factors, hence we will solve it by quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ substituting the values in this equation, we get
$ \Rightarrow t = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 3 \times - 1} }}{{2 \times 3}} = \dfrac{{4 \pm \sqrt {16 + 12} }}{{2 \times 3}} = \dfrac{{4 \pm \sqrt {28} }}{{2 \times 3}}$
$ \Rightarrow t = \dfrac{{4 \pm \sqrt {2 \times 2 \times 7} }}{{2 \times 3}} = \dfrac{{2\left( {2 \pm \sqrt 7 } \right)}}{{2 \times 3}}$
\[ \Rightarrow t = \dfrac{{2 \pm \sqrt 7 }}{3}\]
As $0 < x < \pi $
$ \Rightarrow 0 < \dfrac{x}{2} < \dfrac{\pi }{2}$
So, $\tan \dfrac{x}{2}$ is positive. (In first quadrant tangent is positive)
$\therefore t = \tan \dfrac{x}{2} = \dfrac{{2 + \sqrt 7 }}{3}$
Since, we know that $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$ therefore, $\tan x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 - {{\tan }^2}\dfrac{x}{2}}}$
Now, $\tan x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 - {{\tan }^2}\dfrac{x}{2}}} = \dfrac{{2t}}{{1 - {t^2}}}$
Let us substitute the value of $t$
$ \Rightarrow \tan x = \dfrac{{2\left( {\dfrac{{2 + \sqrt 7 }}{3}} \right)}}{{1 - {{\left( {\dfrac{{2 + \sqrt 7 }}{3}} \right)}^2}}} = \dfrac{{\dfrac{{4 + 2\sqrt 7 }}{3}}}{{1 - {{\dfrac{{\left( {2 + \sqrt 7 } \right)}}{{{3^2}}}}^2}}} = \dfrac{{\dfrac{{4 + 2\sqrt 7 }}{3}}}{{{{\dfrac{{{3^2} - \left( {2 + \sqrt 7 } \right)}}{{{3^2}}}}^2}}}$
It can also be written as,
$ \Rightarrow \tan x = \dfrac{{4 + 2\sqrt 7 }}{3} \times \dfrac{{{3^2}}}{{{3^2} - {{\left( {2 + \sqrt 7 } \right)}^2}}}$
After cancelling out $3$ and expansion of ${\left( {2 + \sqrt 7 } \right)^2}$, we get
$ \Rightarrow \tan x = \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{{3^2} - \left( {4 + 7 + 4\sqrt 7 } \right)}} = \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{9 - \left( {11 + 4\sqrt 7 } \right)}} = \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{9 - 11 - 4\sqrt 7 }}$
\[ \Rightarrow \tan x = \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{ - 2 - 4\sqrt 7 }} = \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{ - \left( {2 + 4\sqrt 7 } \right)}}\]
It can also be written as,
\[ \Rightarrow \tan x = - \dfrac{{3\left( {4 + 2\sqrt 7 } \right)}}{{2 + 4\sqrt 7 }}\]
Take $2$ as a common from both numerator and denominator
\[ \Rightarrow \tan x = - \dfrac{{2 \times 3\left( {2 + \sqrt 7 } \right)}}{{2\left( {1 + 2\sqrt 7 } \right)}}\]
On cancelling $2$, we get
\[ \Rightarrow \tan x = - \dfrac{{3\left( {2 + \sqrt 7 } \right)}}{{1 + 2\sqrt 7 }}\]
Multiplying and dividing $1 - 2\sqrt 7 $ in the above equation, we get
\[ \Rightarrow \tan x = - \dfrac{{3\left( {2 + \sqrt 7 } \right)}}{{1 + 2\sqrt 7 }} \times \dfrac{{1 - 2\sqrt 7 }}{{1 - 2\sqrt 7 }} = - \dfrac{{3\left( {2 - 4\sqrt 7 + \sqrt 7 - 2 \times 7} \right)}}{{{1^2} - {{\left( {2\sqrt 7 } \right)}^2}}}\]
\[ \Rightarrow \tan x = - \dfrac{{3\left( {2 - 3\sqrt 7 - 14} \right)}}{{1 - 28}} = - \dfrac{{3\left( { - 12 - 3\sqrt 7 } \right)}}{{ - 27}} = - \dfrac{{3 \times 3\left( { - 4 - \sqrt 7 } \right)}}{{ - 27}} = - \dfrac{{9\left( { - 4 - \sqrt 7 } \right)}}{{ - 27}}\]
After cancelling out negative signs and division, we get
\[ \Rightarrow \tan x = \dfrac{{\left( { - 4 - \sqrt 7 } \right)}}{3}\]
It can also be written as,
\[ \Rightarrow \tan x = - \left( {\dfrac{{4 + \sqrt 7 }}{3}} \right)\]

Note: One must know how to convert ‘tan’, ‘cot’, ‘sec’, and ‘cosec’ terms in trigonometric formulae for all the terms, especially trigonometric ratios for half-angle, double angle, tripe angles, etc. One should know in which quadrant trigonometric angle of a particular function is positive or negative. We should take care of the calculations so as to be sure of our final answer.