Question

If $\tan \left(\frac{\theta }{2}\right)=t$, then $\frac{1-t^2}{1+t^2}$ is equal to

• A. $\cos \left(\theta \right)$
• B. $\sin \left(\theta \right)$
• C. $\sec \left(\theta \right)$
• D. $\cos \left(2\theta \right)$

Answer 1

The correct option is A

$\cos \left(\theta \right)$

Explanation for correct option

Trigonometry identity:

Given: $\frac{1-t^2}{1+t^2}$

$$\begin{gathered} =\frac{1-{\tan }^2\left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}\left[\because tan\left(\frac{\theta }{2}\right)=t\right] \\ =\frac{1-{\displaystyle \frac{{\sin }^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}}}{1+{\displaystyle \frac{{\sin }^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}}}\mathbf{}\left[\because tan\left(\theta \right)=\frac{sin\left(\theta \right)}{cos\left(\theta \right)}\right] \\ =\frac{\frac{{\cos }^2\left(\frac{\theta }{2}\right)-si{n}^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}}{\frac{{\cos }^2\left(\frac{\theta }{2}\right)+{\sin }^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}} \\ =\frac{{\cos }^2\left(\frac{\theta }{2}\right)-si{n}^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)+{\sin }^2\left(\frac{\theta }{2}\right)}\left[\because co{s}^2\left(x\right)-si{n}^2\left(x\right)=cos\left(2x\right),co{s}^2\left(x\right)+si{n}^2\left(x\right)=1\right] \\ =\cos \left(\theta \right) \end{gathered}$$

Hence, option A is correct.