Question

If ${\sec }^{2}A\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+{\sec }^{2}B\hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+{\sec }^{2}C\hat{k}$, are coplanar then ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$ is

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (D)

$-1$

$\begin{array}{c}\left|\begin{array}{ccc}{\sec }^{2}A& 1& 1\\ 1& {\sec }^{2}B& 1\\ 1& 1& {\sec }^{2}C\end{array}\right|=0\\ \\ C_1\to C_1-C_2{C}_2\to C_2-C_3\\ \\ \left|\begin{array}{ccc}{\tan }^{2}A& 0& 1\\ -{\tan }^{2}B& {\tan }^{2}B& 1\\ 0& -{\tan }^{2}C& {\sec }^{2}C\end{array}\right|=0\\ \\ {\tan }^{2}A\left[{\tan }^{2}B{\sec }^{2}C+{\tan }^{2}C\right]+{\tan }^{2}B{\tan }^{2}C=0\\ \\ \frac{{\sec }^{2}C}{{\tan }^{2}C}+{\cot }^{2}B+{\cot }^{2}A=0\\ \\ \cos e{c}^{2}C+{\cot }^{2}B+{\cot }^{2}A=0\\ \\ {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C=-1\\ \\ Hence,theoptionDisthecorrectanswer.\end{array}$