Question

If $\cos (x-y),\cos x,\cos (x+y)$ are in H.P., where $y\ne 2n\pi ,n\in Z$, then the value of $\left[\cos x\sec \frac{y}{2}\right]$ is/are

(where [.] denotes greatest integer function)

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

Answer: (A), (D)

$1$

$\cos (x-y),\cos x,\cos (x+y)$ are in H.P.

$\Rightarrow \cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x-y)+\cos (x+y)}\Rightarrow \cos x=\frac{2({\cos }^{2}x-{\sin }^{2}y)}{2\cos x\cos y}\Rightarrow {\cos }^{2}x\cos y={\cos }^{2}x-{\sin }^{2}y\Rightarrow {\sin }^{2}y={\cos }^{2}x(1-\cos y)\Rightarrow 1-{\cos }^{2}y={\cos }^{2}x(1-\cos y)\Rightarrow (1-\cos y)(1+\cos y)={\cos }^{2}x(1-\cos y)\Rightarrow 1+\cos y={\cos }^{2}x(\because y\ne 2n\pi \Rightarrow \cos y\ne 1)\Rightarrow {\cos }^{2}x=2{\cos }^2\frac{y}{2}\Rightarrow {\cos }^{2}x{\sec }^2\frac{y}{2}=2\Rightarrow \cos x\sec \frac{y}{2}=\pm \sqrt{2}\Rightarrow [\cos x\times \sec \frac{y}{2}]=\left[\sqrt{2}\right]\text{or}[-\sqrt{2}]\therefore [\cos x\times \sec \frac{y}{2}]=1,-2$