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