Question

If ${\tan }^2\left(\theta \right)=2\tan \left(\varphi \right)+1$, then $\cos \left(2\theta \right)+{\sin }^2\left(\varphi \right)$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. none of these

Answer 1

The correct option is B

$0$

Explanation for correct options

Given: ${\tan }^2\left(\theta \right)=2\tan \left(\varphi \right)+1$ ---------$\left(1\right)$

we know that $\cos \left(2\theta \right)=\frac{1-{\tan }^2\left(\theta \right)}{1+{\tan }^2\left(\theta \right)}$

$$\begin{gathered} \Rightarrow \cos \left(2\theta \right)=\frac{1-\left(2{\tan }^2\left(\varphi \right)+1\right)}{1+\left(2{\tan }^2\left(\varphi \right)+1\right)}\left[\because {\tan }^2\left(\theta \right)=2\tan \left(\varphi \right)+1\right] \\ \Rightarrow \cos \left(2\theta \right)=\frac{1-2{\tan }^2\left(\varphi \right)-1}{1+2{\tan }^2\left(\varphi \right)+1} \\ \Rightarrow \cos \left(2\theta \right)=\frac{-2{\tan }^2\left(\varphi \right)}{2+2{\tan }^2\left(\varphi \right)} \\ \Rightarrow \cos \left(2\theta \right)=\frac{-{\tan }^2\left(\varphi \right)}{{\sec }^2\left(\varphi \right)}\left[1+{\tan }^2\left(x\right)={\sec }^2\left(x\right)\right] \\ \Rightarrow \cos \left(2\theta \right)=\frac{-{\tan }^2\left(\varphi \right)}{{\sec }^2\left(\varphi \right)} \\ \Rightarrow \cos \left(2\theta \right)=\frac{-{\displaystyle \frac{{\sin }^2\left(\varphi \right)}{{\cos }^2\left(\varphi \right)}}}{{\displaystyle \frac{1}{{\cos }^2\left(\varphi \right)}}} \\ \Rightarrow \cos \left(2\theta \right)=-{\sin }^2\left(\varphi \right)------\left(2\right) \end{gathered}$$

From $\left(2\right)$

$$\begin{gathered} \cos \left(2\theta \right)+{\sin }^2\left(\varphi \right)=-{\sin }^2\left(\varphi \right)+{\sin }^2\left(\varphi \right) \\ \Rightarrow =0 \end{gathered}$$

Hence, option $B$ is correct.