HP prob

\hspace{-15}$If $\mathbf{\cos(a-b)\; ,\cos(a)}$ and $\mathbf{\cos(a+b)}$ are in $\textbf{H.P}\;,$ Then\\\\\\ $\mathbf{\displaystyle \cos(a)=\frac{2.\cos(a+b).\cos(a-b)}{\cos(a+b)+\cos(a-b)}.}\;\;$\\\\\\ (Using $\mathbf{\cos(A+B).\cos(A-B)=\cos^2A-\sin^2B}$)\\\\\\ $\mathbf{\cos(a)=\frac{\cos^2(a)-\sin^2(b)}{\cos(a).cos(b)}}$\\\\\\ $\mathbf{\cos^2(a).\cos(b)=\cos^2(a)-\sin^2(b)}$\\\\\\ $\mathbf{\sin^2(b)=\cos^2(a).\left(1-\cos(b)\right)=\cos^2(a).2\sin^2\left(\frac{b}{2}\right)}$\\\\\\ $\mathbf{\sin(b)=\pm\sqrt{2}.\cos(a).\sin\left(\frac{b}{2}\right)}$\\\\\\ $\mathbf{2.\sin\left(\frac{b}{2}\right).\cos\left(\frac{b}{2}\right)=\pm\sqrt{2}.\cos(a).\sin\left(\frac{b}{2}\right)}$\\\\\\ $\boxed{\boxed{\mathbf{\cos(a).\sec\left(\frac{b}{2}\right)=\pm \sqrt{2}}}}$