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

\[\text {Let:} \vec{a} = \left( \sec^2 A \right) \hat {i} + \hat {j} + \hat {k} , \vec{b} = \hat {i} + \left( \sec^2 B \right) \hat {j} + \hat {k} \text{and} \vec{c} = \hat {i} + \hat {j} + \left( \sec^2 C \right) \hat {k} \]

\[\text { We know that three vectors are coplanar iff their scaler triple product is zero . i . e .} , \left[ \vec{a} \vec{b} \vec{c} \right] = 0\]

\[\text { Here,} \left[ \vec{a} \vec{b} \vec{c} \right] = 0\]

\[ \Rightarrow \begin{vmatrix}\sec^2 A & 1 & 1 \\ 1 & \sec^2 B & 1 \\ 1 & 1 & \sec^2 C\end{vmatrix} = 0 \]

\[ \Rightarrow \sec^2 A\left[ \left( \sec^2 B \times \sec^2 C \right) - 1 \right] - 1\left( \sec^2 C - 1 \right) + 1\left( 1 - \sec^2 B \right) = 0\]

\[ \Rightarrow \sec^2 A \sec^2 B \sec^2 C - \sec^2 A - \sec^2 C + 1 + 1 - \sec^2 B = 0\]

\[ \Rightarrow \left( 1 + \tan^2 A \right)\left( 1 + \tan^2 B \right) \left( 1 + \tan^2 C \right) - \left( 1 + \tan^2 A \right) - \left( 1 + \tan^2 C \right) + 1 + 1 - \left( 1 + \tan^2 B \right) = 0\]

\[\Rightarrow 1 + \tan^2 A + \tan^2 B + \tan^2 C + \tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A + \tan^2 A \tan^2 B \tan^2 C - 1 - \tan^2 A - 1 - \tan^2 C + 1 + 1 - 1 - \tan^2 B = 0\]

\[ \Rightarrow \tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A + \tan^2 A \tan^2 B \tan^2 C = 0\]

\[ \Rightarrow \tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A = - \tan^2 A \tan^2 B \tan^2 C\]

\[ \Rightarrow \frac{\tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A}{\tan^2 A \tan^2 B \tan^2 C} = - 1\]

\[ \Rightarrow \cot^2 C + \cot^2 A + \cot^2 B = - 1\]

\[ \Rightarrow {cosec}^2 C - 1 + {cosec}^2 A - 1 + {cosec}^2 B - 1 = - 1\]

\[ \therefore {cosec}^2 A + {cosec}^2 B + {cosec}^2 C = 2\]