Question

If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$ then $(\frac{dy}{dx}{)}_{x=0}$=

• A. $1/2$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

The correct option is D $2$

According to question,

$y=\sqrt{\frac{1-sin2x}{1+sin2x}}$

$⟹y=\sqrt{\frac{si{n}^{2}x+co{s}^{2}x-2cosxsinx}{si{n}^{2}x+co{s}^{2}x+2cosxsinx}}$

$⟹y=\sqrt{\frac{(sinx-cosx{)}^2}{(sinx+cosx{)}^2}}$

$⟹y=\frac{(sinx-cosx)}{(sinx+cosx)}$

$⟹y=\frac{(tanx-1)}{(tanx+1)}$

$⟹y=-tan(\frac{\pi }{4}-x)$

$⟹\frac{dy}{dx}=-se{c}^2(\frac{\pi }{4}-x)\times (-1)$

Putting $x=0$, we get

$\frac{dy}{dx}=2$