The simplified expression of $sin (n + 1) A . sin (n - 1) A + cos (n + 1) A . cos (n - 1) A$ is,

sin 2 A

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cos 2 A

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tan 2 A

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cot 2 A

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Solution

The correct option is B $cos 2 A$
$s i n (n + 1) A . s i n (n - 1) A + c o s (n + 1) A . c o s (n - 1) A$

as $c o s (x - y) = s i n x s i n y + c o s x c o s y$

$\Rightarrow s i n ((n + 1) A) . s i n ((n - 1) A) + c o s ((n - 1) A) . c o s ((n - 1) A)$

$\Rightarrow c o s ((n + 1) - (n - 1) A)$

$\Rightarrow c o s (n + 1 - n + 1) A$

$\Rightarrow c o s 2 A$

option $B$ is correct.

Suggest Corrections

Similar questions

\frac{sin (n + 1) A + 2 sin n A + sin (n - 1) A}{cos (n - 1) A - cos (n + 1) A}

Statement1 : $c o s^{2} A$ + $s i n^{2} A$ = 1

Statement2 : $s e c^{2} A$ + $t a n^{2} A$ = 1

Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=

\frac{{cos}^{2} A + {tan}^{2} A - 1}{{sin}^{2} A} = {tan}^{2} A .

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

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