Determinants Matrices Question 84
Question: In triangle ABC, if $ \begin{vmatrix} 1 & 1 & 1 \\ \cot \frac{A}{2} & \cot \frac{A}{2} & \cot \frac{C}{2} \\ \tan \frac{B}{2}+\tan \frac{C}{2} & \tan \frac{C}{2}+\tan \frac{A}{2} & \tan \frac{A}{2}+\tan \frac{B}{2} \\ \end{vmatrix}=0 $ then the triangle must be
Options:
A) equilateral
B) isosceles
C) obtuse angled
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Applying $ C_1\to C_1-C_2,C_2\to C_2-C_3 $ , we get $ \Delta = \begin{vmatrix} 0 & 0 & 1 \\ \cot \frac{A}{2}-\cot \frac{B}{2} & \cot \frac{B}{2}-\cot \frac{C}{2} & \cot \frac{C}{2} \\ \tan \frac{B}{2}-\tan \frac{A}{2} & \tan \frac{C}{2}-\tan \frac{B}{2} & \tan \frac{A}{2}+\tan \frac{B}{2} \\ \end{vmatrix} $
$ \Delta = \begin{vmatrix} 0 & 0 & 1 \\ \cot \frac{A}{2}-\cot \frac{B}{2} & \cot \frac{B}{2}-\cot \frac{C}{2} & \cot \frac{C}{2} \\ \frac{\cot \frac{A}{2}-\cot \frac{B}{2}}{\cot \frac{A}{2}\cot \frac{B}{2}} & \frac{\cot \frac{B}{2}-\cot \frac{C}{2}}{\cot \frac{B}{2}\cot \frac{C}{2}} & \tan \frac{A}{2}+\tan \frac{B}{2} \\ \end{vmatrix} $
$ =( \cot \frac{A}{2}-\cot \frac{B}{2} )( \cot \frac{B}{2}-\cot \frac{C}{2} ) $
$ \times \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & \cot \frac{C}{2} \\ \tan \frac{A}{2}\tan \frac{B}{2} & \tan \frac{B}{2}\tan \frac{C}{2} & \tan \frac{A}{2}\tan \frac{B}{2} \\ \end{vmatrix} $
$ =( \cot \frac{A}{2}-\cot \frac{B}{2} )( \cot \frac{B}{2}-\cot \frac{C}{2} ) $
$ ( \tan \frac{C}{2}-\tan \frac{A}{2} )\tan \frac{B}{2} $ Since $ \Delta =0 $ , we have $ \cot \frac{A}{2}=\cot \frac{B}{2} $ Or $ \cot \frac{B}{2}=\cot \frac{C}{2} $
$ \tan \frac{A}{2}=\tan \frac{C}{2} $ Hence, the triangle is definitely isosceles.
BETA
