First slide

Multiple and sub- multiple Angles

Question

If $|\begin{matrix} A & D & 1 \\ B & E & 1 \\ C & F & 1 \end{matrix}| = |\begin{matrix} π & \frac{π}{2} & 3 \\ B - C & E - F & 0 \\ C & F & 1 \end{matrix}|$ then
Statement 1: $\tan A + \tan B + \tan C = \tan A . \tan B . \tan C$
Statement 2: $\cot D + \cot E + \cot F = \cot D . \cot E . \cot F$ Then

Easy

A

only one is true
B

only 2 is true
C

both 1 and 2 are true
D

None of them is true

Solution

If $A + B + C = π$
$\begin{array}{l} \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \cdot \tan C \\ If D + E + F = \frac{π}{2} then \\ \cot D + \cot E + \cot F = \cot D \cdot \cot E \cdot \cot F \end{array}$
Now,
$|\begin{matrix} A & D & 1 \\ B & E & 1 \\ C & F & 1 \end{matrix}| (a p p l y R_{1} \to R_{1} + R_{2} + R_{3}, R_{2} \to R_{2} - R_{3}) = |\begin{matrix} A + B + C & D + E + F & 3 \\ B - C & E - F & 0 \\ C & F & 1 \end{matrix}| = |\begin{matrix} π & \frac{π}{2} & 3 \\ B - C & E - F & 0 \\ C & F & 1 \end{matrix}| (g i v e n)$
$\Rightarrow A + B + C = π and D + E + F = \frac{π}{2} ∴ Statement (1) and statement (2) both are true .$

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