Question

If ${\tan }^{2}A=2{\tan }^{2}B+1$, then $\cos 2A+{\sin }^{2}B$ equals

• A. -1
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (C)

Given : ${\tan }^{2}A=2{\tan }^{2}B+1$
$\Rightarrow 1+{\tan }^{2}A=2{\tan }^{2}B+1+1$
$\Rightarrow {\sec }^{2}A=2{\sec }^{2}B$
$\Rightarrow {\cos }^{2}B=2{\cos }^{2}A$
$\Rightarrow {\cos }^{2}B=1+\cos 2A$
$\Rightarrow {\cos }^{2}B-1=\cos 2A$
$\Rightarrow -{\sin }^{2}B=\cos 2A$
$\Rightarrow \cos 2A+{\sin }^{2}B=0$