Question

If $\cos \theta =\frac{2\sqrt{ab}}{a+b}$, find the value $\tan \theta$.

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $\frac{a-b}{2\sqrt{ab}}$
• D. $\frac{(a+b{)}^2}{(a-b{)}^2}$

Answer 1

The correct option is C $\frac{a-b}{2\sqrt{ab}}$

$\sin \theta =\sqrt{1-(\cos \theta {)}^2}=\sqrt{1-\frac{4ab}{(a+b{)}^2}}=\sqrt{\frac{(a+b{)}^2-4ab}{(a+b{)}^2}}=\frac{a-b}{a+b}$

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{a-b}{a+b}}{\frac{2\sqrt{ab}}{a+b}}=\frac{a-b}{2\sqrt{ab}}$

Therefore, Answer is $\frac{a-b}{2\sqrt{ab}}$