Question

If $\tan x=b/a$, then
$\sqrt{(a+b)(a-b)}+\sqrt{(a-b)(a+b)}$ is equal to

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

Answer 1

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

Ans.(b). We have

using rationlization we get,

$\sqrt{\left(\frac{a+b}{a-b}\right)}+\sqrt{\left(\frac{a-b}{a+b}\right)}=\frac{a+b+a-b}{\surd (a^2-b^2)}$

$=\frac{2a}{\surd (a^2-b^2)}=\frac{2}{\surd \{1-{\left(b/a\right)}^2\}}$

$=\frac{2}{\surd (1-{\tan }^{2}x)}=\frac{2\cos x}{\surd ({\cos }^{2}x-{\sin }^{2}x)}=\frac{2\cos x}{\surd \left(\cos 2x\right)}$