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