Question

If $\frac{b}{a}=\tan x$ then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ is equal to

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

Answer 1

The correct option is B $\frac{2\cos x}{\sqrt{\cos 2x}}$

$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{2a}{\sqrt{a^2-b^2}}$
$=\frac{2}{\sqrt{1-{\left(\frac{b}{a}\right)}^2}}=\frac{2}{\sqrt{1-{\tan }^{2}x}}$
$=\frac{2\cos x}{\sqrt{{\cos }^{2}x-{\sin }^{2}x}}=\frac{2\cos x}{\sqrt{\cos 2x}}$
$\therefore$ Choice (b) is correct.