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If $\cos x = \dfrac{1}{2}$, the value of the expression $\dfrac{\cos 6x+6 \cos 4x+15 \cos 2x +10}{\cos 5x+5 \cos 3x +10 \cos x}$  is

1. $\frac{1}{2}$
2. $1$
3. $\frac{1}{4}$
4. $0$

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Answer: $B$

Solution:

Let $$y = \frac{\cos6x + 6\cos4x + 15\cos2x + 10}{\cos5x + 5\cos3x + 10\cos x}\tag{i}$$

$6\cos4x$ and $15\cos2x$ can be written as:

$$6\cos4x = \cos x + 5\cos4x; \tag{ii}$$

$$15\cos2x = 10\cos2x + 5\cos2x\tag{iii}$$

Now, we know that:

$$2\cos a \cos b = \cos\frac{(a+b)}{2} \ cos \frac{a-b}{2}\tag{iv}$$

Using $(i), \;(ii)\;and\;(iii)$, we get:

$$\implies y = \frac{\underbrace{\cos6x +( \cos4x} + \underbrace{5\cos4x)+(5\cos2x} + \underbrace{10\cos2x) +10}}{\cos5x + 5\cos3x + 10\cos x}\tag{v}$$

Using $(iv)$ and $(v)$, we get:

$$\implies y = \frac{2\cos5x\cos x + 10\cos3x\cos x + 10(\cos2x + \underbrace{1}_\text {= \cos 0})}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x(\cos5x + 5\cos3x) + 10(\cos2x + \underbrace{\cos 0})}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x(\cos 5x + 5\cos3x) +10\times2\cos x \cos x}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x [\cos5 + 5\cos3x + 10\cos x]}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = 2\cos x$$

Now, substitute $x = \frac{1}{2}$, we get:

$$y = 2\times \frac{1}{2} = 1$$

Answer: $\therefore B$ is the correct option.

by Boss (19.3k points)
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