Question

If $\cos (x-y), \cos x, \cos (x +y)$ are three distinct numbers which are in harmonic progression and $\cos x \neq \cos y$, 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, $\cos x=\frac{2 \cos (x-y) \cos (x +y)}{\cos (x +y)+\cos (x-y)}$
$\cos x=\frac{\cos 2 x+\cos 2 y}{2 \cos x \cdot \cos y}$
$\cos x=\frac{2 \cos ^{2} x+2 \cos ^{2} y-2}{2 \cos x \cdot \cos y}$
$\cos ^{2} x \cdot \cos y=\cos ^{2} x+\cos ^{2} y-1$
$\cos ^{2} x(\cos y-1)=\left(\cos ^{2} y-1\right)$
$\cos ^{2} x(1-\cos y)=\left(1-\cos ^{2} y\right)$
$\cos ^{2} x\left(2 \sin ^{2} \frac{y}{2}\right)=\sin ^{2} y$
$\cos ^{2} x\left(2 \sin ^{2} \frac{y}{2}\right)=\left(2 \sin \frac{y}{2} \cdot \cos \frac{y}{2}\right)^{2}$
$\cos ^{2} x\left(2 \sin ^{2} \frac{y}{2}\right)=4 \sin ^{2} \frac{y}{2} \cdot \cos ^{2} \frac{y}{2}$
$\cos ^{2} x=\left(2 \cos ^{2} \frac{y}{2}-1\right)+1$
$\cos y+1=\cos ^{2} x$
or $1+\cos y=\cos ^{2} x$