Question

If $\cos \left( {x - y} \right),\cos x$ and $\cos \left( {x + y} \right)$ are in harmonic progression, then the value of $\cos x\sec \left( {\frac{y}{2}} \right)$ is:
(A) $\pm \sqrt 2$ (B) $\pm \sqrt 3$ (C) $\pm 2$ (D) $\pm 1$

Answer 1

Hint: Use the formula for harmonic mean between two numbers and then simplify the expression.

According to the information given in the question, $\cos \left( {x - y} \right),\cos x$ and $\cos \left( {x + y} \right)$ are in harmonic progression.
We know that, if three numbers a, b and c are in H.P. then b is the harmonic mean of a and c its value is:
$\Rightarrow b = \frac{{2ac}}{{a + c}}$
Using this result for $\cos \left( {x - y} \right),\cos x$ and $\cos \left( {x + y} \right)$ , we’ll get:
$\Rightarrow \cos x = \frac{{2\cos \left( {x - y} \right)\cos \left( {x + y} \right)}}{{\cos \left( {x - y} \right) + \cos \left( {x + y} \right)}} .....(i)$
And we also know that, $2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$ . Using this result for the above expression, we’ll get:
$\Rightarrow \cos x = \frac{{\left( {\cos 2x + \cos 2y} \right)}}{{2\cos x\cos y}}, \\\ \Rightarrow 2{\cos ^2}x\cos y = \cos 2x + \cos 2y \\\$
Using $\cos 2x = 2{\cos ^2}x - 1$ , we’ll get:

$$\Rightarrow 2{\cos ^2}x\cos y = 2{\cos ^2}x - 1 + 2{\cos ^2}y - 1, \\\ \Rightarrow 2{\cos ^2}x\cos y - 2{\cos ^2}x = 2{\cos ^2}y - 2, \\\ \Rightarrow 2{\cos ^2}x\left( {\cos y - 1} \right) = 2\left( {{{\cos }^2}y - 1} \right), \\\ \Rightarrow {\cos ^2}x\left( {\cos y - 1} \right) = \left( {\cos y + 1} \right)\left( {\cos y - 1} \right), \\\ \Rightarrow {\cos ^2}x = \cos y + 1, \\\ \Rightarrow {\cos ^2}x = 2{\cos ^2}\left( {\frac{y}{2}} \right) - 1 + 1, \\\$$

$\Rightarrow {\cos ^2}x = 2{\cos ^2}\left( {\frac{y}{2}} \right),$

$$\Rightarrow \frac{{{{\cos }^2}x}}{{{{\cos }^2}\left( {\frac{y}{2}} \right)}} = 2, \\\ \Rightarrow {\cos ^2}x{\sec ^2}\left( {\frac{y}{2}} \right) = 2, \\\ \Rightarrow \cos x\sec \left( {\frac{y}{2}} \right) = \pm \sqrt 2 \\\$$

Therefore, the value of $\cos x\sec \left( {\frac{y}{2}} \right)$ is $\pm \sqrt 2$ . Thus (A) is correct option.
Note:
Harmonic mean between two numbers a and c is always given as:
H.M. $= \frac{{2ac}}{{a + c}}$.
But in this case, three numbers a, b and c are in harmonic progression, therefore b is the harmonic mean of a and c. So, we get:
H.M. $= b$
$\Rightarrow b = \frac{{2ac}}{{a + c}}$
This is what we used in the above question.