Question

If \[\tan \dfrac{{B - C}}{2} = x\cot \dfrac{A}{2}\], then find the value of x.
A. \[\dfrac{{c - a}}{{c + a}}\]
B. \[\dfrac{{a - b}}{{a + b}}\]
C. \[\dfrac{{b - c}}{{b + c}}\]
D. None of these

Answer 1

Hint: First we will multiply \[\cot \dfrac{A}{2}\tan \dfrac{A}{2}\] with the left side of the equation. Then simplify the left-side expression by using the difference between the two sine functions. Then apply sine law to solve the equation and we get the value of x.

Formula used:
In \[\Delta ABC\], \[A + B + C = \pi \]
Complementary angle of trigonometry
\[\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \theta \]
\[\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\]
\[\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]
Sine law
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Complete step by step solution:
Given equation is \[\tan \dfrac{{B - C}}{2} = x\cot \dfrac{A}{2}\]
Multiply \[\cot \dfrac{A}{2}\tan \dfrac{A}{2}\] with the left side of the equation
\[ \Rightarrow \tan \dfrac{{B - C}}{2}\cot \dfrac{A}{2}\tan \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
Substitute \[A = \pi - B - C\].
\[ \Rightarrow \tan \dfrac{{B - C}}{2}\tan \dfrac{{\pi - B - C}}{2}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
Apply the formula \[\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \theta \]
\[ \Rightarrow \tan \dfrac{{B - C}}{2}\cot \dfrac{{B + C}}{2}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
Now apply the formula \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] and \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[ \Rightarrow \dfrac{{\sin \dfrac{{B - C}}{2}}}{{\cos \dfrac{{B - C}}{2}}} \cdot \dfrac{{\cos \dfrac{{B + C}}{2}}}{{\sin \dfrac{{B + C}}{2}}}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
\[ \Rightarrow \dfrac{{2\sin \dfrac{{B - C}}{2}\cos \dfrac{{B + C}}{2}}}{{2\cos \dfrac{{B - C}}{2}\sin \dfrac{{B + C}}{2}}}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
Now applying \[\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\] and \[\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\]
\[ \Rightarrow \dfrac{{\sin B - \sin C}}{{\sin B + \sin C}}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\] …(i)
Sine law is
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {say} \right)\]
\[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\]
Putting \[\sin C = ck\], \[\sin B = bk\] in equation (i)
\[ \Rightarrow \dfrac{{bk - ck}}{{bk + ck}}\cot \dfrac{A}{2} = x\cot \dfrac{A}{2}\]
Cancel out \[\cot \dfrac{A}{2}\] from both sides
\[ \Rightarrow \dfrac{{k\left( {b - c} \right)}}{{k\left( {b + c} \right)}} = x\]
\[ \Rightarrow \dfrac{{\left( {b - c} \right)}}{{\left( {b + c} \right)}} = x\]
Thus the value of x is \[\dfrac{{b - c}}{{b + c}}\].
Hence option C is the correct option.

Note: Students often confused with the formulas of sum of two sine function and difference of two sine function. They used \[\sin A + \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\] and \[\sin A - \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]. The correct formulas are \[\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\] and \[\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\].