Question

In a $\Delta ABC$, $\tan A$ and$\tan B$ satisfy the inequation $\sqrt 3 {x^2} - 4x + \sqrt 3 < 0$, then
A.${a^2} + {b^2} + ab > {c^2}$
B.${a^2} + {b^2} - ab < {c^2}$
C.${a^2} + {b^2} > {c^2}$
D.None of these

Answer 1

Hint : In this problem, we need to solve the given inequality satisfied by $\tan A$ and $\tan B$ in $\Delta ABC$First, we use factorization method for finding the value of$x.$Here, We use the quadratic formula is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. then, comparing with the formula $A + B + C = \pi$ and also substitute all the values in to this equation, $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$.
An inequation is a statement that an inequality or a non-equality holds between two values.

Complete step by step solution:
In the given problem,
Let $\tan A$and$\tan B$satisfies the inequation
$\sqrt 3 {x^2} - 4x + \sqrt 3 < 0$ ----------(1)
By using the quadratic equation for factoring the equation (1), we can get
The quadratic formula is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
From the equation (1), we have $a = \sqrt 3 ,b = - 4,c = \sqrt 3$
$x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(\sqrt 3 )(\sqrt 3 )} }}{{2\sqrt 3 }}$
$x = \dfrac{{4 \pm \sqrt {16 - 4(3)} }}{{2\sqrt 3 }}$, since ${(\sqrt 3 )^2} = 3$
On further simplification, then
$x = \dfrac{{4 \pm \sqrt {16 - 12} }}{{2\sqrt 3 }} = \dfrac{{4 \pm \sqrt 4 }}{{2\sqrt 3 }}$
By solving the square root of the numerator, we get
$x = \dfrac{{4 \pm 2}}{{2\sqrt 3 }}$
Now, We have to find the value of ‘x’, then
If $x = \dfrac{{4 + 2}}{{2\sqrt 3 }}$, then
$\Rightarrow x = \dfrac{6}{{2\sqrt 3 }} = \dfrac{3}{{\sqrt 3 }} = \sqrt 3$
If $x = \dfrac{{4 - 2}}{{2\sqrt 3 }}$, then
$\Rightarrow x = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$
Therefore, $x \in \left( {\sqrt 3 ,\dfrac{1}{{\sqrt 3 }}} \right)$
The factors of $\sqrt 3 {x^2} - 4x + \sqrt 3 < 0$ is $(x - \sqrt 3 )(\sqrt 3 x - 1) < 0$
Therefore,$\dfrac{1}{{\sqrt 3 }} < x < \sqrt 3$
Let us assume, tan A and tan B satisfies ‘x’ value, then
Since, $\dfrac{1}{{\sqrt 3 }} < x < \sqrt 3$
$\Rightarrow \dfrac{1}{{\sqrt 3 }} < \tan A < \sqrt 3$ or $\dfrac{1}{{\sqrt 3 }} < \tan B < \sqrt 3$
Now, expanding the equation of tan (x) as ${\tan ^{ - 1}}x$, we get
$\Rightarrow {\tan ^{ - 1}}(\dfrac{1}{{\sqrt 3 }}) < A < {\tan ^{ - 1}}(\sqrt 3 )$ or ${\tan ^{ - 1}}(\dfrac{1}{{\sqrt 3 }}) < B < {\tan ^{ - 1}}(\sqrt 3 )$ ------(2)
We know that, from the trigonometric degree table, ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {30^ \circ },{\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = {60^ \circ }$.
Now, we have to draw a triangle ABC as given below.

From the equation (2), we need to finding the radian of ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{6},{\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{\pi }{3}$.
$\Rightarrow \dfrac{\pi }{6} < A < \dfrac{\pi }{3}$or$\dfrac{\pi }{6} < B < \dfrac{\pi }{3}$
By adding$A$and$B$, we can get
$\Rightarrow \dfrac{\pi }{6} + \dfrac{\pi }{6} < A + B < \dfrac{\pi }{3} + \dfrac{\pi }{3}$
$\Rightarrow \dfrac{{2\pi }}{6} < A + B < \dfrac{{2\pi }}{3}$
$\Rightarrow \dfrac{\pi }{3} < A + B < \dfrac{{2\pi }}{3}$
Now, on comparing the formula $A + B + C = \pi$, then the value of$A + B = \pi - C$
$\Rightarrow \dfrac{\pi }{3} < \pi - C < \dfrac{{2\pi }}{3}$
\Rightarrow - \pi + \dfrac{\pi }{3} < \- C < \- \pi + \dfrac{{2\pi }}{3}$$$$$$ By solving radian of both sides of the equation, we get \Rightarrow - \dfrac{{2\pi }}{3} < - C < - \dfrac{\pi }{3}$$ \Rightarrow \dfrac{{2\pi }}{3} > C > \dfrac{\pi }{3}$Since,$C > \dfrac{\pi }{3}$, then$\cos C > \cos \dfrac{\pi }{3}$We know that from the trigonometric table$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$. Therefore, The value of$\cos \dfrac{\pi }{3}$is$\dfrac{1}{2}$, then we can get We use the$\cos C < \dfrac{1}{2}$substitute into the formula we know the formula is$\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$, then we can get$ \Rightarrow \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} < \dfrac{1}{2}$$ \Rightarrow {a^2} + {b^2} - {c^2} < \dfrac{{2ab}}{2}$On further simplification, we can expanding the ‘ab’ to LHS we can get$ \Rightarrow {a^2} + {b^2} - ab < {c^2}$Since,$C < \dfrac{{2\pi }}{3}$,$\cos C < \cos \dfrac{{2\pi }}{3}$By using the ASTC rule of trigonometry, the angle$\pi - \dfrac{\pi }{3}$or angle$180 - \theta $lies in the second quadrant. cosine function are negative here, hence the angle must in negative, then$ \Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$$ \Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}$The value of$\cos \dfrac{{2\pi }}{3}$is$ - \dfrac{1}{2}$Therefore,$\cos C > \dfrac{{ - 1}}{2}$We use the formula is$\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$, then we can get$\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} > \dfrac{{ - 1}}{2}$${a^2} + {b^2} - {c^2} > \dfrac{{ - 2ab}}{2}$Bu cancel the 2 from numerator and denominator, then${a^2} + {b^2} + ab > {c^2}$Therefore, the final answer is option (A)${a^2} + {b^2} + ab > {c^2}$$.
So, the correct answer is “Option A”.

Note : Simply this can also be solve by using a ASTC rule i.e.,
$\Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$
By using the ASTC rule of trigonometry, the angle $\pi - \dfrac{\pi }{3}$ or angle $180 - \theta$ lies in the second quadrant. cosine function are negative here, hence the angle must in negative, then
$\Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$
$\Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}$
While solving this type of question, we must know about the ASTC rule.
And also know the cosine sum or difference identity, for this we have a standard formula. To find the value for the trigonometry function we need the table of trigonometry ratios for standard angles