Question

If the vectors $\overrightarrow{r_1}=({\sec }^{2}A,1,1)$ , $\overrightarrow{r_2}=(1,{\sec }^{2}B,1)$ ,$\overrightarrow{r_3}=(1,1,{\sec }^{2}C)$ are coplanar, then ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$

• A. Is equal to $0$
• B. Is equal to $1$
• C. Is equal to $-1$
• D. Is greater than $2$

Answer 1

The correct option is C Is equal to $-1$

$\left|\begin{array}{ccc}{\sec }^{2}A& 1& 1\\ 1& {\sec }^{2}B& 1\\ 1& 1& {\sec }^{2}C\end{array}\right|=0$
${\sec }^{2}A({\sec }^{2}B{\sec }^{2}C-1)-1({\sec }^{2}C-1)+(1-{\sec }^{2}B)$
${\sec }^{2}A{\sec }^{2}B{\sec }^{2}C-{\sec }^{2}A-{\tan }^{2}C-{\tan }^{2}B=0$
${\sec }^{2}A\left(\right(1+{\tan }^{2}B\left)\right(1+{\tan }^{2}C)-1)-{\tan }^{2}C-{\tan }^{2}B=0$
${\sec }^{2}A({\tan }^{2}B{\tan }^{2}C+{\tan }^{2}B+{\tan }^{2}C)-{\tan }^{2}C-{\tan }^{2}B=0$
${\sec }^{2}A{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A+{\tan }^{2}B{\tan }^{2}A=0$
$(1+{\tan }^{2}A)\left({\tan }^{2}B{\tan }^{2}C\right)+{\tan }^{2}C{\tan }^{2}A+{\tan }^{2}B{\tan }^{2}A=0$
${\tan }^{2}B{\tan }^{2}C+{\tan }^{2}A{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A+{\tan }^{2}B{\tan }^{2}A=0$
Dividing throughout by ${\tan }^{2}A{\tan }^{2}B{\tan }^{2}C$, we get
${\cot }^{2}A+1+{\cot }^{2}B+{\cot }^{2}C=0$
${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C=-1$