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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

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Question : 151 of 160

Marks: +1, -0

\text{ If } \begin{vmatrix} \cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{vmatrix}=0

B

$(2n+1) \frac{\pi}{2}$
$(2n+1) \pi$
$n \pi$
$2n \pi$

Solution:

We have,

\begin{vmatrix} \cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{vmatrix}=0

\Rightarrow \cos(A+B)[\cos A \cos B - \sin A \sin B]

+ \sin(A+B)[\sin A \cos B + \cos A \sin B]

+ \cos 2B [\sin^2 A + \cos^2 A] = 0

\Rightarrow \cos(A+B) \cos(A+B)

+ \sin(A+B) \sin(A+B) + \cos 2B = 0

\Rightarrow \cos^2(A+B) + \sin^2(A+B) + \cos 2B = 0

\Rightarrow 1 + \cos 2B = 0

\Rightarrow 1 + 2\cos^2 B - 1 = 0

\Rightarrow \cos^2 B = 0 \Rightarrow \cos^2 B = \cos^2 \frac{\pi}{2}

\Rightarrow B = n\pi \pm \frac{\pi}{2} = \frac{\pi}{2}(2n \pm 1)

\therefore B = \frac{\pi}{2}(2n+1)

$ $

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