Question

If ${\sin }^{2}x+{\cos }^{2}y=2{\sec }^{2}z$, then
(where $m,n,t\in Z$)

• A. $x=(2m+1)\frac{\pi }{2}$
• B. $y=n\pi$
• C. $z=t\pi$
• D. $x=y+z$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x=(2m+1)\frac{\pi }{2}$
B $y=n\pi$
C $z=t\pi$

LHS $={\sin }^{2}x+{\cos }^{2}y\le 2$
RHS $=2{\sec }^{2}z\ge 2$
Hence LHS $=$ RHS only when
${\sin }^{2}x=1,{\cos }^{2}y=1$ and ${\sec }^{2}z=1$
$\Rightarrow {\cos }^{2}x=0,{\sin }^{2}y=0$ and ${\cos }^{2}z=1$
$\Rightarrow \cos x=0,\sin y=0$ and $\sin z=0$
$\Rightarrow x=(2m+1)\frac{\pi }{2},y=n\pi ,z=t\pi$
where $m,n,t\in Z$

Now, assuming $x=y+z$
$(2m+1)\frac{\pi }{2}=(n+t)\pi \Rightarrow (2m+1)=2(n+t)$
We know that,
$2(n+t)\to \text{even number}2m+1\to \text{odd number}$
So,
$x\ne y+z$