Question

If $\sin \left(\theta \right)=\frac{2t}{1+t^2}$ and $\left(\theta \right)$ lies in the second quadrant , then $\cos \left(\theta \right)$ is equals to

• A. $\left(\frac{1-t^2}{1+t^2}\right)$
• B. $\left(\frac{t^2-1}{1+t^2}\right)$
• C. $\frac{-\left|1-t^2\right|}{\left(1+t^2\right)}$
• D. $\frac{\left(1+t^2\right)}{\left|1-t^2\right|}$

Answer 1

The correct option is C

$\frac{-\left|1-t^2\right|}{\left(1+t^2\right)}$

Explanation for the correct option

Step 1: Given expression

$\sin \left(\theta \right)=\frac{2t}{1+t^2}$ and $\left(\theta \right)$ lies in the second quadrant

Step 2: Simplification and calculate the value of $\cos \left(\theta \right)$

From the given

$\therefore \cos \left(\theta \right)<0$

So value of $\cos \left(\theta \right)$ is negative

Step 3: Apply the formula of $\cos \left(\theta \right)$

$\begin{array}{rcl}& \Rightarrow & \cos \left(\theta \right)=-\sqrt{1-{\sin }^2\left(\theta \right)}\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{1-{\left(\frac{2t}{1+t^2}\right)}^2}\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{1-\frac{4t^2}{{\left(1+t^2\right)}^2}}\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{1-\frac{4t^2}{1+t^2+2t^2}}\left[\therefore {\left(a-b\right)}^2=a^2+b^2+2ab\right]\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{\frac{1+t^2+2t^2-4t^2}{1+t^2+2t^2}}\left(takeL.C.M\right)\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{\frac{1+t^2-2t^2}{1+t^2+2t^2}}\\ & \Rightarrow & \cos \left(\theta \right)=-\sqrt{\frac{{\left(1-t^2\right)}^2}{{\left(1+t^2\right)}^2}}\\ & \Rightarrow & \cos \left(\theta \right)=-\frac{\left|1-t^2\right|}{\left(1+t^2\right)}\end{array}$

The value of $\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left(\theta \right)\end{array}=-\frac{\left|1-t^2\right|}{\left(1+t^2\right)}$

Hence , option $\left(\mathit{C}\right)$ is correct.