Question

If $\left|tanA\right|<1,$ and $\left|A\right|$ is acute then $\frac{\sqrt{1+sin2A}+\sqrt{1-sin2A}}{\sqrt{1+sin2A}-\sqrt{1-sin2A}}$ is equal to

• A. tan A
• B. $-$tan A
• C. cot A
• D. $-$cot A

Answer 1

Answer: (C)

cot A

Let, $y=\frac{\sqrt{1+sin2A}+\sqrt{1-sin2A}}{\sqrt{1+sin2A}-\sqrt{1-sin2A}}$
$=\frac{\sqrt{si{n}^{2}A+co{s}^{2}A+2sinA.cosA}+\sqrt{si{n}^{2}A+co{s}^{2}A-2sinA.cosA}}{\sqrt{si{n}^{2}A+co{s}^{2}A+2sinA.cosA}-\sqrt{si{n}^{2}A+co{s}^{2}A-2sinA.cosA}}$
$=\frac{\sqrt{(cosA+sinA{)}^2}+\sqrt{(cosA-sinA{)}^2}}{\sqrt{(cosA+sinA{)}^2}-\sqrt{(cosA-sinA{)}^2}}$
Given $\mid tanA\mid \le 1\Rightarrow \mid cosA\mid \ge \mid sinA\mid$,
since $\mid A\mid$ is acute $\Rightarrow cosA\ge \mid sinA\mid$
therefore,
$y=\frac{(cosA+sinA)+(cosA-sinA)}{(cosA+sinA)-(cosA-sinA)}=cotA$

Hence, option 'C' is correct.