Question

If $\frac{{\cos }^2(x+y)-{\cos }^2(x-y)}{-4\sin x\sin y}=\underset{x\to 0^{+}}{lim}\text{sgn(sgn(sgn}x\left)\right)$, where $\text{sgn}$ is signum function, then which of the following statements is/are correct?

• A. $x=n\pi$ and $y=m\pi$, where $n,m$ are both even integers
• B. $x=n\pi$ and $y=m\pi$, where $n$ is even and $m$ is an odd integer.
• C. $x=n\pi$ and $y=m\pi$, where $n,m$ are both odd integers
• D. $x=n\pi$ and $y=m\pi$, where $n$ is odd and $m$ is an even integer.

Answer 1

Answer: (A), (C)

The correct options are
A $x=n\pi$ and $y=m\pi$, where $n,m$ are both even integers
C $x=n\pi$ and $y=m\pi$, where $n,m$ are both odd integers

$\frac{{\cos }^2(x+y)-{\cos }^2(x-y)}{-4\sin x\sin y}=\underset{x\to 0^{+}}{lim}\text{sgn(sgn(sgn}x\left)\right)$

$\Rightarrow \frac{{\cos }^2(x+y)-{\cos }^2(x-y)}{2\left(\cos \right(x+y)-\cos (x-y\left)\right)}=1$

$\Rightarrow \frac{\cos (x+y)+\cos (x-y)}{2}=1$

$\Rightarrow \cos x\cos y=1$

Now we know that,
$-1\le \cos x\le 1$, and $-1\le \cos y\le 1$
So, $\cos x=1$ and $\cos y=1$
or $\cos x=-1$ and $\cos y=-1$
$\Rightarrow x=2n\pi$ and $y=2m\pi$ $(n,m\in Z)$
or $x=(2n+1)\pi$ and $y=(2m+1)\pi$ $(n,m\in Z)$

$x=n\pi$ and $y=m\pi$ where $n,m$ are both even integers or both odd integers.