# 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$

Last updated date: 23rd Mar 2023

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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.

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.

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