Question

If $\theta$ is an acute angle and $\sin \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}},$ then $\tan \theta$ is equal to

• A. $x^2–1$
• B. $\sqrt{x^2-1}$
• C. $\sqrt{x^2+1}$
• D. $x^2+1$

Answer 1

The correct option is B

$\sqrt{x^2-1}$

Explanation for the correct option.

Step 1. Find the value of $\cos \frac{\theta }{2}$

Let's consider the given functions

$\sin \frac{\theta }{2}=\sqrt{\frac{(x-1)}{2x}}$

Squaring on both sides,

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

Step 2. Find the value of $\sin \theta$.

Multiply by $\sin \frac{\theta }{2}$ and $\cos \frac{\theta }{2}$

$$\begin{gathered} \sin \frac{\theta }{2}\cos \frac{\theta }{2}=\sqrt{\frac{(x–1)}{2x}}\times \sqrt{\frac{(x+1)}{2x}} \\ \Rightarrow \sin \frac{\theta }{2}\cos \frac{\theta }{2}=\frac{\sqrt{\left(x-1\right)\left(x+1\right)}}{2x} \\ \Rightarrow 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\frac{\sqrt{x^2-1}}{x} \\ \Rightarrow \sin \theta =\frac{\sqrt{x^2-1}}{x} \end{gathered}$$

Step 3. Find the value of $\cos \theta$:

Squaring both sides $\sin \theta =\frac{\sqrt{x^2-1}}{x}$ and use the identity to find the value of $\cos \theta$.

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

Step 4. Find the value of $\tan \theta$.

Using $\tan \theta =\frac{\sin \theta }{\cos \theta }$, the value is found as:

$\begin{array}{rcl}\tan \theta & =& \frac{\sin \theta }{\cos \theta }\\ & =& \frac{{\displaystyle \frac{\sqrt{x^2-1}}{x}}}{{\displaystyle \frac{1}{x}}}\\ & =& \sqrt{x^2-1}\end{array}$

Hence, the correct option is B.