Question

If ${\sin }^{2}x+{\cos }^{2}y=2{\sec }^{2}z$ then y is

• A. $n\pi$
• B. $2n\pi$
• C. $\frac{n\pi }{2}$
• D. $(n+\frac{1}{2})\pi$

Answer 1

The correct option is A $n\pi$

We have $si{n}^{2}x+co{s}^{2}y=2se{c}^{2}z$

$L.H.S=si{n}^{2}x+co{s}^{2}y\le 2$

$R.H.S=2se{c}^{2}z\ge 2$

Hence $L.H.S=R.H.S$ only when $si{n}^{2}x=1,co{s}^{2}y=1,se{c}^{2}z=1$

$co{s}^{2}x=0,si{n}^{2}y=0,co{s}^{2}z=1\Rightarrow cosx=0,siny=0,sinz=1$

$\Rightarrow x=(2m+1)\frac{\pi }{2},y=n\pi ,z=t\pi$ where $m,n.t$ are integerss.