Question

If $\cos (x-y),\cos x,\cos (x+y)$ are three distinct numbers which are in harmonic progression and $\cos x\ne \cos y$, then $1+\cos y$ is equal to

• A. ${\cos }^{2}x$
• B. $-{\cos }^{2}x$
• C. ${\cos }^{2}x-1$
• D. ${\cos }^{2}x-2$

Answer 1

Answer: (A)

$\cos (x-y),\cos x,\cos (x+y)$ are in HP.
Then, $\cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x+y)+\cos (x-y)}$
$\cos x=\frac{\cos 2x+\cos 2y}{2\cos x\cdot \cos y}$
$\cos x=\frac{2{\cos }^{2}x+2{\cos }^{2}y-2}{2\cos x\cdot \cos y}$
${\cos }^{2}x\cdot \cos y={\cos }^{2}x+{\cos }^{2}y-1$
${\cos }^{2}x(\cos y-1)=\left({\cos }^{2}y-1\right)$
${\cos }^{2}x(1-\cos y)=\left(1-{\cos }^{2}y\right)$
${\cos }^{2}x\left(2{\sin }^2\frac{y}{2}\right)={\sin }^{2}y$
${\cos }^{2}x\left(2{\sin }^2\frac{y}{2}\right)={\left(2\sin \frac{y}{2}\cdot \cos \frac{y}{2}\right)}^2$
${\cos }^{2}x\left(2{\sin }^2\frac{y}{2}\right)=4{\sin }^2\frac{y}{2}\cdot {\cos }^2\frac{y}{2}$
${\cos }^{2}x=\left(2{\cos }^2\frac{y}{2}-1\right)+1$
$\cos y+1={\cos }^{2}x$
or $1+\cos y={\cos }^{2}x$