Question

In a {{\Delta ABC,}}\;{\text{if}}\;\left| {\begin{array}{*{20}{c}} {{\text{cot}}\dfrac{{\text{A}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ {\text{1}}&{\text{1}}&{\text{1}} \end{array}} \right| = 0, then the triangle must be
A. Equilateral
B. Isosceles
C. Right angled
D. Scalene

Answer 1

Firstly, simplify the determinant by making elements of a row zero as much as possible and then open the determinant and solve it. Solve it using trigonometric relations, and you will finally get some condition between the angles, through that condition choose the answer from given options.

Complete step by step answer:
In order to find if the given triangle ABC is either equilateral or isosceles or right angled or scalene, we will use the given information in the question that is
\left| {\begin{array}{*{20}{c}} {{\text{cot}}\dfrac{{\text{A}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ {\text{1}}&{\text{1}}&{\text{1}} \end{array}} \right| = 0
Simplifying this determinant by applying column operation as follows
Operating ${{\text{C}}_{\text{1}}} - {{\text{C}}_{\text{2}}}\;{\text{and}}\;{{\text{C}}_{\text{2}}} - {{\text{C}}_{\text{3}}}$ we will get

\Rightarrow \left| {\begin{array}{*{20}{c}} {{\text{cot}}\dfrac{{\text{A}}}{{\text{2}}} - {\text{cot}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ {{\text{1}} - 1}&{{\text{1}} - 1}&{\text{1}} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {{\text{cot}}\dfrac{{\text{A}}}{{\text{2}}} - {\text{cot}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ 0&0&{\text{1}} \end{array}} \right| = 0 \\\

Now from the trigonometric relation of tangent and cotangent, we know that they are multiplicative inverse of each other, that is $\cot x = \dfrac{1}{{\tan x}}$
Using this to convert cotangents present in the equation into tangent, we will get

\Rightarrow \left| {\begin{array}{*{20}{c}} {\dfrac{1}{{\tan \dfrac{{\text{A}}}{{\text{2}}}}} - \dfrac{1}{{\tan \dfrac{{\text{B}}}{{\text{2}}}}}}&{\dfrac{1}{{\tan \dfrac{{\text{B}}}{{\text{2}}}}} - \dfrac{1}{{\tan \dfrac{{\text{C}}}{{\text{2}}}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ 0&0&{\text{1}} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {\dfrac{{\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}}}{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}}}}&{\dfrac{{\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{B}}}{{\text{2}}}}}{{\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ {{\text{tan}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}}&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ 0&0&{\text{1}} \end{array}} \right| = 0 \\\

Now taking ${\text{tan}}\dfrac{{\text{B}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}\;{\text{and}}\;{\text{tan}}\dfrac{{\text{C}}}{{\text{2}}} - {\text{tan}}\dfrac{{\text{B}}}{{\text{2}}}$ common from column one and column two respectively, we will get

{\dfrac{1}{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}}}}&{\dfrac{1}{{\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}}&{{\text{cot}}\dfrac{{\text{C}}}{{\text{2}}}} \\\ 1&1&{{\text{tan}}\dfrac{{\text{A}}}{{\text{2}}}{\text{ + tan}}\dfrac{{\text{B}}}{{\text{2}}}} \\\ 0&0&{\text{1}} \end{array}} \right| = 0$$ Now, expanding the determinant across row three, we will get $$ \Rightarrow \left( {\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right)\left( {\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{B}}}{{\text{2}}}} \right)\left( {\dfrac{1}{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}}} - \dfrac{1}{{\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}} \right) = 0$$ Simplifying it further we will get

\Rightarrow \left( {\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right)\left( {\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{B}}}{{\text{2}}}} \right)\left( {\dfrac{{\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}}} - \dfrac{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}}}{{\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}} \right) = 0 \\
\Rightarrow \left( {\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right)\left( {\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{B}}}{{\text{2}}}} \right)\left( {\dfrac{{\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}}}{{\tan \dfrac{{\text{A}}}{{\text{2}}}\tan \dfrac{{\text{B}}}{{\text{2}}}\tan \dfrac{{\text{C}}}{{\text{2}}}}}} \right) = 0 \\
\Rightarrow \left( {\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right)\left( {\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{B}}}{{\text{2}}}} \right)\left( {\tan \dfrac{{\text{C}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right) = 0 \\

$$Now, if we separately equate the above three terms to zero we will get$$

\Rightarrow \left( {\tan \dfrac{{\text{B}}}{{\text{2}}} - \tan \dfrac{{\text{A}}}{{\text{2}}}} \right) = 0 \\
\Rightarrow \tan \dfrac{{\text{B}}}{{\text{2}}} = \tan \dfrac{{\text{A}}}{{\text{2}}} \\
\Rightarrow \dfrac{{\text{B}}}{{\text{2}}} = \dfrac{{\text{A}}}{{\text{2}}} \\
\Rightarrow {\text{B}} = {\text{A}} \\

Similarly, ${\text{B}} = {\text{C}}\;{\text{or}}\;{\text{C}} = {\text{A}}$ So, there is a chance of holding true for all three conditions, but we are sure that either of the three possibilities is true, that is either of the two angles of the triangle are equal. **$\therefore $ Option B is correct.** **Note:** In order to simplify the matrix or determinant, always approach the matrix either via row or column. Using column and row both can lead you to lengthy and wrong solutions. Always try to make all the elements zero of a row or column.