Question

If in any $\Delta ABC \cot \dfrac{A}{2},\cot \dfrac{B}{2},\cos \dfrac{C}{2}$ are in A. P. then [MP PET 2003]
A. $\cot \dfrac{A}{2}\cot \dfrac{B}{2}=4$
B. $\cot \dfrac{A}{2}\cot \dfrac{C}{2}=3$
C. $\cot \dfrac{B}{2}\cot \dfrac{C}{2}=1$
D. $\cot \dfrac{B}{2}\tan \dfrac{C}{2}=0$

Answer 1

Hint:In the question we have given a triangle with $\cot \dfrac{A}{2},\cot \dfrac{B}{2},\cot \dfrac{C}{2}$ in A.P. which means that A, B, and C are also in A.P. Use the given relation and the property of the triangle, you will get an expression. You can simplify it further either by changing it into tan of angle or by using trigonometric identities.

Formula used: The common difference of an AP can be calculated as: $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = ... = a_{n}- a$.


Complete step by step solution: Given in triangle, $\cot \dfrac{A}{2},\cot \dfrac{B}{2},\cot \dfrac{C}{2}$ are in A. P.
$A+B+C=180$
$ \cot \dfrac{A}{2}+\cot \dfrac{C}{2}=2 \cot \dfrac{B}{2}$
$= \cot (\dfrac{A}{2})+\cot (\dfrac{C}{2})=2 \cot (\dfrac{180-(A+C)}{2})$
$=\dfrac{1}{\tan (\dfrac{A}{2})}+\dfrac{1}{\tan (\dfrac{C}{2})}=2 \dfrac{1}{\tan (\dfrac{A+C}{2})}$
$=\dfrac{\tan (\dfrac{C}{2}+\tan (\dfrac{A}{2}}{\tan (\dfrac{C}{2} \times \tan (\dfrac{A}{2}}=2 \tan \dfrac{A+C}{2}$
$=\dfrac{ \tan (\dfrac{A}{2})+\tan (\dfrac{C}{2})}{1- \tan (\dfrac{A}{2}) \tan (\dfrac{C}{2})}$
$=1- \tan (\dfrac{A}{2}) \tan (\dfrac{C}{2})=2 \tan {A}{2} \tan {C}{2}$
$=1=3 \tan \dfrac{A}{2} \tan \dfrac{C}{2}$
Hence,
$\cot \dfrac{A}{2} \cot{C}{2}=3$


Thus, Option (B) is correct.

Note: The law of cotangents in trigonometry describes the relationship between a triangle's side lengths and the cotangents of its three-angle halves. As the $\cot \dfrac{A}{2},\cot \dfrac{B}{2},\cot \dfrac{C}{2}$ are in A.P, then A,B and C are also in A.P. Remember the relation $a+c=2b$ is obtained from the common difference of A.P.