Question

$x=co{\sec }^2\theta ,y={\sec }^2\theta ,z=\frac{1}{1-{\sin }^2\theta {\cos }^2\theta }$

$xyz$ is equal to

• A. $x+y+z$
• B. $xy+z$
• C. $x+y-z$
• D. $\frac{x+y}{z}$

Answer 1

Answer: (A), (B)

The correct options are
A $x+y+z$
B $xy+z$

$z=\frac{1}{1-{\sin }^2\theta {\cos }^2\theta }=\frac{{co\sec }^2\theta {\sec }^2\theta }{{co\sec }^2\theta {\sec }^2\theta -1}=\frac{xy}{xy-1}$ ...(1)

$xy={co\sec }^2\theta {\sec }^2\theta =\frac{1}{{\sin }^2\theta {\cos }^2\theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }$

$=\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }={co\sec }^2\theta {+\sec }^2\theta$ ...(2)

From (1) and (2), we get

$xyz-z=xy\Rightarrow xyz=x+y+z$