If $c o s (x - y), c o s x, c o s (x + y)$ are three distinct numbers which are harmonic progression and $c o s x \neq c o s y$ then $1 + c o s y =$

$c o s^{2} x$

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$- c o s^{2} x$

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$c o s^{2} x - 1$

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$c o s^{2} x - 2$

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Solution

The correct option is A
$c o s^{2} x$

$b = \frac{2 a c}{a + c} \Rightarrow c o s x = \frac{2 c o s (x - y) c o s (x + y)}{c o s (x - y) + c o s (x + y)} \Rightarrow 1 + c o s y = c o s^{2} x$

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Similar questions

If $c o s (x - y), c o s x, c o s (x + y)$ are three distinct numbers which are harmonic progression and $c o s x \neq c o s y$ then $1 + c o s y =$

x, y, z

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & cos (x - y) & cos (x - z) cos (y - x) & 1 & cos (y - z) cos (z - x) & cos (z - y) & 1 \end{matrix} ⎤ ⎥ ⎦

A

\frac{1}{3}

\frac{1}{4}

- π \leq x \leq π

- π \leq y \leq π

cos x + cos y = 2

cos (x - y)

y = {cos}^{- 1} cos x,

\frac{d y}{d x}

x = \frac{5 π}{4}

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