Question

If $\tan \left(x\right)=\frac{b}{a}$, then $\sqrt{\frac{(a+b)}{(a-b)}}+\sqrt{\frac{(a-b)}{(a+b)}}=?$

• A. $\frac{2\sin \left(x\right)}{\sqrt{\sin \left(2x\right)}}$
• B. $\frac{2\cos \left(x\right)}{\sqrt{\cos \left(2x\right)}}$
• C. $\frac{2\cos \left(x\right)}{\sqrt{\sin \left(2x\right)}}$
• D. $\frac{2\sin \left(x\right)}{\sqrt{\cos \left(2x\right)}}$

Answer 1

The correct option is B

$\frac{2\cos \left(x\right)}{\sqrt{\cos \left(2x\right)}}$

Explanation for correct option

Given $\tan \left(x\right)=\frac{b}{a}$

$$\begin{gathered} \sqrt{\frac{(a+b)}{(a-b)}}+\sqrt{\frac{(a-b)}{(a+b)}} \\ =\frac{(a+b)+(a-b)}{\sqrt{(a-b)(a+b)}} \\ =\frac{2a}{\sqrt{a^2-b^2}}\left[\because a^2-b^2=(a-b)(a+b)\right] \\ =\frac{2a}{a\sqrt{1-{\left({\displaystyle \frac{b}{a}}\right)}^2}} \\ =\frac{2}{\sqrt{1-{\tan }^2\left(x\right)}}\left[\because \tan \left(x\right)=\frac{b}{a}\right] \\ =\frac{2}{\sqrt{1-{\displaystyle \frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}}}}\left[\because \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\right] \\ =\frac{2\cos \left(x\right)}{\sqrt{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)}} \\ =\frac{2\cos \left(x\right)}{\sqrt{\cos \left(2x\right)}}\left[\because \cos \left(2x\right)={\cos }^2\left(x\right)-{\sin }^2\left(x\right)\right] \end{gathered}$$

Hence option (B) is correct i.e. $\frac{2\cos \left(x\right)}{\sqrt{\cos \left(2x\right)}}$