Question

If $\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$, then the value of $\tan A$ is

• A. $\frac{x^2+y^2}{x^2-y^2}$
• B. $\frac{2xy}{x^2+y^2}$
• C. $\frac{2xy}{x^2-y^2}$
• D. $\frac{2xy}{y^2-x^2}$

Answer 1

The correct option is C

$\frac{2xy}{x^2-y^2}$

Explanation for the correct option:

Step 1: Solving the given function

$\begin{array}{l}1+\cos A=2{\cos }^2\left(\frac{A}{2}\right).......\left(i\right)\\ 1-\cos A=2{\sin }^2\left(\frac{A}{2}\right).......\left(ii\right)\end{array}$

Divide $\left(i\right)$ by $\left(ii\right)$ and take square roots on both the sides, we get

$\begin{array}{c}\therefore \sqrt{\frac{1+\cos A}{1-\cos A}}=\sqrt{\frac{2{\cos }^2\left(\frac{A}{2}\right)}{2{\sin }^2\left(\frac{A}{2}\right)}}\\ =\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}\\ =\cot \left(\frac{A}{2}\right)\end{array}$

Given that $\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$,

Therefore,

$\begin{array}{c}\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{x}{y}\\ \Rightarrow \tan \left(\frac{A}{2}\right)=\frac{y}{x}\end{array}$

Step 2: Apply half angle formula to find the function.

We know $\tan A=\frac{2\tan \left(\frac{A}{2}\right)}{1-{\tan }^2\left(\frac{A}{2}\right)}$

Put $\tan \left(\frac{{\displaystyle A}}{{\displaystyle 2}}\right)=\frac{y}{x}$, Which implies

$\begin{array}{c}\therefore \tan A=\frac{2\left(\frac{y}{x}\right)}{1-{\left(\frac{y}{x}\right)}^2}\\ =\frac{2\left(\frac{y}{x}\right)}{\frac{x^2-y^2}{x^2}}\\ =\frac{2xy}{x^2-y^2}\end{array}$

Hence, the correct option is (C).