Question

If the vectors (sec² A) $\hat{i}+\hat{j}+\hat{k},\hat{i}+\left({\sec }^{2}B\right)\hat{j}+\hat{k},\hat{i}+\hat{j}+\left({\sec }^{2}C\right)\hat{k}$ are coplanar, then find the value of cosec² A + cosec² B + cosec² C.

Answer 1

$$\begin{gathered} \mathrm{Let}:\overrightarrow{a}=\left({\sec }^{2}A\right)\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+\left({\sec }^{2}B\right)\hat{j}+\hat{k}\mathrm{and}\overrightarrow{c}=\hat{i}+\hat{j}+\left({\sec }^{2}C\right)\hat{k} \\ \mathrm{W}e\mathrm{know}\mathrm{that}\mathrm{three}\mathrm{vectors}\mathrm{are}\mathrm{coplanar}\mathrm{iff}\mathrm{their}\mathrm{scaler}\mathrm{triple}\mathrm{product}\mathrm{is}\mathrm{zero}.\mathrm{i}.\mathrm{e}.,\left[\overrightarrow{a}\mathit{}\overrightarrow{b}\mathit{}\overrightarrow{c}\right]=0 \\ \mathrm{Here},\left[\overrightarrow{\mathrm{a}}\mathit{}\overrightarrow{\mathrm{b}}\mathit{}\overrightarrow{\mathrm{c}}\right]=0 \\ \Rightarrow \left|\begin{array}{ccc}{\sec }^{2}A& 1& 1\\ 1& {\sec }^{2}B& 1\\ 1& 1& {\sec }^{2}C\end{array}\right|=0 \\ \Rightarrow {\sec }^{2}A\left[\left({\sec }^{2}B\times {\sec }^{2}C\right)-1\right]-1\left({\sec }^{2}C-1\right)+1\left(1-{\sec }^{2}B\right)=0 \\ \Rightarrow {\sec }^{2}A{\sec }^{2}B{\sec }^{2}C-{\sec }^{2}A-{\sec }^{2}C+1+1-{\sec }^{2}B=0 \\ \Rightarrow \left(1+{\tan }^{2}A\right)\left(1+{\tan }^{2}B\right)\left(1+{\tan }^{2}C\right)-\left(1+{\tan }^{2}A\right)-\left(1+{\tan }^{2}C\right)+1+1-\left(1+{\tan }^{2}B\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow 1+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C+{\tan }^{2}A{\tan }^{2}B+{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A+{\tan }^{2}A{\tan }^{2}B{\tan }^{2}C-1-{\tan }^{2}A-1-{\tan }^{2}C+1+1-1-{\tan }^{2}B=0 \\ \Rightarrow {\tan }^{2}A{\tan }^{2}B+{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A+{\tan }^{2}A{\tan }^{2}B{\tan }^{2}C=0 \\ \Rightarrow {\tan }^{2}A{\tan }^{2}B+{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A=-{\tan }^{2}A{\tan }^{2}B{\tan }^{2}C \\ \Rightarrow \frac{{\tan }^{2}A{\tan }^{2}B+{\tan }^{2}B{\tan }^{2}C+{\tan }^{2}C{\tan }^{2}A}{{\tan }^{2}A{\tan }^{2}B{\tan }^{2}C}=-1 \\ \Rightarrow {\cot }^{2}C+{\cot }^{2}A+{\cot }^{2}B=-1 \\ \Rightarrow {\mathrm{cosec}}^{2}C-1+{\mathrm{cosec}}^{2}A-1+{\mathrm{cosec}}^{2}B-1=-1 \\ \therefore {\mathrm{cosec}}^{2}A+{\mathrm{cosec}}^{2}B+{\mathrm{cosec}}^{2}C=2 \end{gathered}$$