$\frac{sin (n + 1) A + 2 sin n A + sin (n - 1) A}{cos (n - 1) A - cos (n + 1) A}$ is equal to

tan \frac{A}{2}

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cot \frac{A}{2}

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tan A

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cot A

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Solution

The correct option is C $cot \frac{A}{2}$
$sin (n + 1) A + 2 sin n A + sin (n - 1) A = sin (n A + A) + sin (n A - A) + 2 sin n A$
$= 2 sin n A cos A + 2 sin n A$ $(∵ sin (a + b) + sin (a - b) = 2 sin a cos b)$
$= 2 sin n A (1 + cos A)$
$= 4 sin n A {cos}^{2} \frac{A}{2}$ $(∵ 1 + cos 2 A = 2 {cos}^{2} A)$

$cos (n - 1) A - cos (n + 1) A = cos (n A - A) - cos (n A + A)$
$= 2 sin n A sin A$ $(∵ cos (a - b) - cos (a + b) = 2 sin a sin b)$
$= 4 sin n A sin \frac{A}{2} cos \frac{A}{2}$ $(∵ sin 2 A = 2 sin A cos A)$

$\frac{sin (n + 1) A + 2 sin n A + sin (n - 1) A}{cos (n - 1) A - cos (n + 1) A} = \frac{4 sin n A {cos}^{2} \frac{A}{2}}{4 sin n A sin \frac{A}{2} cos \frac{A}{2}} = \frac{cos \frac{A}{2}}{sin \frac{A}{2}} = cot \frac{A}{2}$

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Similar questions

sin (n + 1) A . sin (n - 1) A + cos (n + 1) A . cos (n - 1) A

(sec A + tan A - 1) (sec A - tan A + 1) =

\frac{1}{(sec A + tan A)} - \frac{1}{cos A} = \frac{1}{cos A} - \frac{1}{(sec A - tan A)}

$\frac{(sec A + tan A - 1) (sec A - tan A + 1)}{tan A} =$

\frac{1}{\sec A + \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A - \tan A}

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