Question

If $x=co{\sec }^2\theta ,y={\sec }^2\theta ,z=\frac{1}{1-{\sin }^2\theta {\cos }^2\theta }$ then $ta{n}^{2}2\theta$ is equal to

• A. $\frac{4(z+1)y^2}{z(2-y{)}^2}$
• B. $\frac{4(z-1)y^2}{z(2-y{)}^2}$
• C. $\frac{4(z-1)x^2}{z(2-x{)}^2}$
• D. $\frac{4x^2(y-1)}{y^2(2x-1)}$

Answer 1

Answer: (B), (C)

$\frac{4(z-1)y^2}{z(2-y{)}^2}$
$\frac{4(z-1)x^2}{z(2-x{)}^2}$

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

$\Rightarrow si{n}^2\theta =\frac{1}{x},{\cos }^2\theta =\frac{1}{y},1-{\sin }^2\theta {\cos }^2\theta =\frac{1}{z}$

$si{n}^{2}2\theta =2{sin}^2\theta {cos}^2\theta =4(1-\frac{1}{z})=\frac{4(z-1)}{z}$ ...(1)

$cos2\theta =2co{s}^2\theta -1=\frac{2-y}{y}$ ...(2)

$cos2\theta =1-2si{n}^2\theta =\frac{x-2}{x}$ ...(3)

From (1) and (2)

$ta{n}^{2}2\theta =\frac{4(z-1)y^2}{z{(2-y)}^2}$

And from (1) and (3)

$ta{n}^{2}2\theta =\frac{4(z-1)x^2}{z(x-2{)}^2}$