Question

If $\sec \left(\theta \right)=1\frac{1}{4}$, then $\tan \left(\frac{{\displaystyle \theta }}{{\displaystyle 2}}\right)=$

• A. $\frac{1}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is A

$\frac{1}{3}$

Step 1 : Apply the formula $\cos \left(A\right)=\frac{1-{\tan }^2\frac{A}{2}}{1+{\tan }^2\frac{A}{2}}$

$\begin{array}{l}\therefore \sec \theta =1\frac{1}{4}\\ \Rightarrow \frac{1}{\cos \theta }=\frac{5}{4}\\ \Rightarrow \cos \theta =\frac{4}{5}\\ \Rightarrow \frac{4}{5}=\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\end{array}$

Step 2 : Use the cross multiply method

$\begin{array}{rcl}4\times \left[1+{\tan }^2\left(\frac{\theta }{2}\right)\right]& =& 5\times \left[1-{\tan }^2\left(\frac{\theta }{2}\right)\right]\\ 4+4{\tan }^2\left(\frac{\theta }{2}\right)& =& 5-5{\tan }^2\left(\frac{\theta }{2}\right)\\ 4{\tan }^2\left(\frac{\theta }{2}\right)+5{\tan }^2\left(\frac{\theta }{2}\right)& =& 5-4\\ 9{\tan }^2\left(\frac{\theta }{2}\right)& =& 1\\ {\tan }^2\left(\frac{\theta }{2}\right)& =& \frac{1}{9}\end{array}$

Step 3: Take the square root

$\begin{array}{rcl}\therefore {\tan }^2\left(\frac{\theta }{2}\right)& =& \frac{1}{9}\\ & =& \sqrt{\frac{1}{9}}\\ & =& \frac{1}{3}\end{array}$

Hence , the correct option is $\left(A\right)$.