If cos(x-y),cos(x),cos(x+y) are in H.P then the value of cox(x)sec(y/2) is ?

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Answer 1 · 2018-03-22T02:50:42

Because they are in Harmonic Progress (H.P.) we have

$\frac{1}{cos (x + y)} + \frac{1}{cos (x - y)} = \frac{2}{cos x}$ $1/cos(x+y)+1/cos(x-y)=2/cosx$

Hence

$cos x \cdot [cos (x + y) + cos (x - y)] = 2 \cdot cos (x + y) \cdot cos (x - y)$ $cosx*[cos(x+y)+cos(x-y)]=2*cos(x+y)*cos(x-y)$

${cos}^{2} x \cdot cos y = {cos}^{2} x - {sin}^{2} y$ $cos^2x*cosy=cos^2x-sin^2y$

${cos}^{2} x = cos y + 1$ $cos^2x=cosy+1$

$cos² x = 2 cos² (y/2)$

$(cos²x) / (cos²(y/2)) = 2$

$cos² x. sec² (y/2) = 2$

Taking absolute values we have

$|cosx*sec(y/2)|=sqrt2$

or

$cosx*sec(y/2)=+-sqrt2$