
In any triangle \[ABC,\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}} = \]
A. \[\frac{{a - b}}{{a + b}}\]
В. \[\frac{{a - b}}{c}\]
C. \[\frac{{a - b}}{{a + b + c}}\]
D. \[\frac{c}{{a + b}}\]
Answer
161.1k+ views
Hint:
We must determine the value of \[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\] and as a result, we shall employ numerous trigonometric identities, such as \[\tan = \frac{{\sin }}{{\cos }}\] then, after simplifying it using the identity \[\sin (x - y) = \sin x\cos y - \sin y\cos x\] we shall multiply by \[2\sin \left( {\frac{C}{2}} \right)\] and we will next utilize the previously described identity to simplify and obtain the appropriate value of \[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\]
Formula used:
Value of tan function is
\[\tan = \frac{{\sin }}{{\cos }}\]
Complete step-by step solution:
We have been provided in the question that there is a triangle \[ABC\]
And we are to find the value of
\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\]
The formula that can be used to calculate the value of the tan function for any angle in terms of sin and cos is \[\tan = \frac{{\sin }}{{\cos }}\]
Now, we have to simplify the above expression by using trigonometry identity \[\tan = \frac{{\sin }}{{\cos }}\] we get
\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}} = \frac{{\frac{{\sin A/2}}{{\cos A/2}} - \frac{{\sin B/2}}{{\cos B/2}}}}{{\frac{{\sin A/2}}{{\cos A/2}} + \frac{{\sin B/2}}{{\cos B/2}}}}\]
Now, we have to solve the numerator and denominator to make to less complicated, we get\[ = \frac{{\sin \frac{A}{2} - \cos \frac{B}{2} - \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}{{\sin \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}\]
On further simplification of numerator since \[\sin (x - y) = \sin x\cos y - \sin y\cos x\]it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}\]
On further simplification of denominator since \[\sin (x + y) = \sin x\cos y + \sin y\cos x\] it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin \left( {\frac{{A + B}}{2}} \right)}}\]
Now, we have to apply triangle’s sum of angles property, we have
\[A + B + C = {180^{^0}}\]
The above can be written as below,
\[\left( {\frac{{A + B}}{2}} \right) = {90^0} - \frac{C}{2}\]
Now, on simplifying further using trigonometry identity.
Since we already know that \[\sin ({90^0} - \theta ) = \cos \theta \] it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\left( {\sin {{90}^0} - \frac{C}{2}} \right)}} = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\cos \left( {\frac{C}{2}} \right)}}\]------- (1)
Now, we have to multiply the equation (1) by \[2\sin \left( {\frac{C}{2}} \right)\] we have
\[\frac{{2\sin \left( {\frac{C}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)}}{{2\sin \left( {\frac{C}{2}} \right)\cos \frac{C}{2}}}\]
After multiplication, since \[\sin 2\theta = 2\sin \theta \cos \theta \] it becomes
\[ \frac{{2\sin \left( {{{90}^\circ } - \frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin 2\left( {\frac{C}{2}} \right)}}\]
On further simplification since \[\sin ({90^0} - \theta ) = \cos \theta \] we get
\[ = \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}}}{{\sin C}}\]
Since \[\sin x - \sin y = 2\cos \frac{{x + y}}{2}\sin \frac{{x - y}}{2}\] the equation obtained above becomes,
\[ = \frac{{\sin A - \sin B}}{{\sin C}}\]---- (2)
We have been already know that the sine rule is
\[\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = k\]
So, now the equation (2) will become
\[\frac{{\frac{a}{k} - \frac{b}{k}}}{{\frac{c}{k}}}\]
On further simplification, it becomes
\[ = \frac{{a - b}}{c}\]
Therefore, in any triangle \[ABC,\]\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}} = \frac{{a - b}}{c}\]
Hence, the option B is correct
NOTE:
In such cases where we need to apply trigonometric identities to solve the problem, we must pay close attention to our fundamentals, which means knowing all of the identities as well as when and where to use them. Such questions can be complex, and you may become perplexed as to how to begin, especially if the question includes an initial condition. When employing the trigonometric identity, you may make a mistake.
We must determine the value of \[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\] and as a result, we shall employ numerous trigonometric identities, such as \[\tan = \frac{{\sin }}{{\cos }}\] then, after simplifying it using the identity \[\sin (x - y) = \sin x\cos y - \sin y\cos x\] we shall multiply by \[2\sin \left( {\frac{C}{2}} \right)\] and we will next utilize the previously described identity to simplify and obtain the appropriate value of \[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\]
Formula used:
Value of tan function is
\[\tan = \frac{{\sin }}{{\cos }}\]
Complete step-by step solution:
We have been provided in the question that there is a triangle \[ABC\]
And we are to find the value of
\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}}\]
The formula that can be used to calculate the value of the tan function for any angle in terms of sin and cos is \[\tan = \frac{{\sin }}{{\cos }}\]
Now, we have to simplify the above expression by using trigonometry identity \[\tan = \frac{{\sin }}{{\cos }}\] we get
\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}} = \frac{{\frac{{\sin A/2}}{{\cos A/2}} - \frac{{\sin B/2}}{{\cos B/2}}}}{{\frac{{\sin A/2}}{{\cos A/2}} + \frac{{\sin B/2}}{{\cos B/2}}}}\]
Now, we have to solve the numerator and denominator to make to less complicated, we get\[ = \frac{{\sin \frac{A}{2} - \cos \frac{B}{2} - \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}{{\sin \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}\]
On further simplification of numerator since \[\sin (x - y) = \sin x\cos y - \sin y\cos x\]it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{B}{2} \cdot \cos \frac{B}{2}}}\]
On further simplification of denominator since \[\sin (x + y) = \sin x\cos y + \sin y\cos x\] it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin \left( {\frac{{A + B}}{2}} \right)}}\]
Now, we have to apply triangle’s sum of angles property, we have
\[A + B + C = {180^{^0}}\]
The above can be written as below,
\[\left( {\frac{{A + B}}{2}} \right) = {90^0} - \frac{C}{2}\]
Now, on simplifying further using trigonometry identity.
Since we already know that \[\sin ({90^0} - \theta ) = \cos \theta \] it becomes
\[ = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\left( {\sin {{90}^0} - \frac{C}{2}} \right)}} = \frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\cos \left( {\frac{C}{2}} \right)}}\]------- (1)
Now, we have to multiply the equation (1) by \[2\sin \left( {\frac{C}{2}} \right)\] we have
\[\frac{{2\sin \left( {\frac{C}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)}}{{2\sin \left( {\frac{C}{2}} \right)\cos \frac{C}{2}}}\]
After multiplication, since \[\sin 2\theta = 2\sin \theta \cos \theta \] it becomes
\[ \frac{{2\sin \left( {{{90}^\circ } - \frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)}}{{\sin 2\left( {\frac{C}{2}} \right)}}\]
On further simplification since \[\sin ({90^0} - \theta ) = \cos \theta \] we get
\[ = \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}}}{{\sin C}}\]
Since \[\sin x - \sin y = 2\cos \frac{{x + y}}{2}\sin \frac{{x - y}}{2}\] the equation obtained above becomes,
\[ = \frac{{\sin A - \sin B}}{{\sin C}}\]---- (2)
We have been already know that the sine rule is
\[\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = k\]
So, now the equation (2) will become
\[\frac{{\frac{a}{k} - \frac{b}{k}}}{{\frac{c}{k}}}\]
On further simplification, it becomes
\[ = \frac{{a - b}}{c}\]
Therefore, in any triangle \[ABC,\]\[\frac{{\tan \frac{A}{2} - \tan \frac{B}{2}}}{{\tan \frac{A}{2} + \tan \frac{B}{2}}} = \frac{{a - b}}{c}\]
Hence, the option B is correct
NOTE:
In such cases where we need to apply trigonometric identities to solve the problem, we must pay close attention to our fundamentals, which means knowing all of the identities as well as when and where to use them. Such questions can be complex, and you may become perplexed as to how to begin, especially if the question includes an initial condition. When employing the trigonometric identity, you may make a mistake.
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