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Special Integrals -4

Find a cos ...

Question

Find a $\frac{cos 3 x}{sin 2 x sin 4 x} + \frac{cos 5 x}{sin 4 x sin 6 x} + \frac{cos 7 x}{sin 6 x sin 8 x} + \frac{cos 9 x}{sin 8 x sin 10 x} = \frac{1}{2} (c o s e c x) [c o s e c 2 x - c o s e c a x]$

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Solution

Let $f (x) = \frac{cos 3 x}{sin 2 x sin 4 x} + \frac{cos 5 x}{sin 4 x sin 6 x} + \frac{cos 7 x}{sin 6 x sin 8 x} + \frac{cos 9 x}{sin 8 x sin 10 x}$
Multiply and divide by $(2 sin x)$ in whole expression

$f (x) = \frac{sin 4 x - sin 2 x}{2 sin x sin 2 x sin 4 x} + \frac{sin 6 x - sin 4 x}{2 sin x sin 4 x sin 6 x} + \frac{sin 8 x - sin 6 x}{2 sin x sin 6 x sin 8 x} + \frac{sin 10 x - sin 8 x}{2 sin x sin 8 x sin 10 x}$
$= \frac{c o s e c x [c o s e c 2 x - c o s e c 10 x]}{2}$
Ans: a = 10

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