First slide

Multiple and sub- multiple Angles

Question

If A+B+C=0, then
The value of $|\begin{matrix} A & \tan A & c o r A \cot B \\ B & \tan B & \cot B \cot C \\ C & \tan C & \cot C \cot A \end{matrix}| = |\begin{matrix} λ & \tan A \tan B \tan C & 1 \\ - A & \tan B + \tan C & \cot C (\cot B + \cot A) \\ C & \tan C & \cot C \cot A \end{matrix}|$

Moderate

A

0
B

1
C

2
D

-1

Solution

$\begin{array}{l} A + B + C = 0 \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C and \\ \cot A \cot B + \cot B \cot C + \cot C \cot A = 1 \\ N o w, |\begin{matrix} A & \tan A & c o t A c o t B \\ B & \tan B & c o t B c o t C \\ C & \tan C & c o t C c o t A \end{matrix}| \\ R_{1} \to R_{1} + R_{2} + R_{3}, R_{2} \to R_{2} + R_{3} \\ = |\begin{matrix} A + B + C & \tan A + \tan B + \tan C & c o t A c o t B + c o t B c o t C + c o t C c o t A \\ B + C & \tan B + \tan C & c o t B c o t C + c o t C c o t A \\ C & \tan C & c o t C c o t A \end{matrix}| \\ = |\begin{matrix} 0 & \tan A \tan B \tan C & 1 \\ - A & \tan B + \tan C & c o t C (c o t B + c o t A) \\ C & \tan C & c o t C c o t A \end{matrix}| \\ \Rightarrow λ = 0 \end{array}$

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