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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$

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Last updated date: 17th Jul 2024
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Answer
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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.