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If $\cos a=k\cos(b/2)$ and $\sec(a+b)\sec(a-b)=2\sec a$, find the value(s) of $k$.

I have converted in $\sec$'s into $\cos$ and then applied the formulae of $\cos(a+b)\cos (a-b)=\cos^2 a-\sin^2 b$ and simplified further but it did not lead to my answer

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  • $\begingroup$ Please use MathJax to type the equations. $\endgroup$ – Matti P. May 29 '18 at 6:33
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Some trig/algebraic simplifications should follow after your first step:

$c$ is abbreviated for $\cos ...$

$$ \frac{c_a}{c_{b/2}} = k$$

$$ \frac{2}{c_a}= \frac{1}{c_a^2-c_b^2} $$

let $x= c_{b/2} , Q= 2 x^2,\, P=2(2x^2-1)^2$ and simplify

$$ k^2 Q-kx-P=0$$

A quadratic in $k$ having 2 values in terms of trig function of $b$

$$ k= \frac{x\pm \sqrt{x^2+4PQ}}{2Q} $$

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