Question

If $y=\ell n\sqrt{\frac{1-\sin x}{1+\sin x}}$, then $\frac{dy}{dx}$ equals

• A. $\sec x$
• B. $-secx$
• C. $cosecx$
• D. $\sec x\tan x$

Answer 1

Answer: (B)

$-secx$

$y=ln\sqrt{\frac{1-sinx}{1+sinx}}$

$y=\frac{1}{2}ln\frac{1-sinx}{1+sinx}$

$y=\frac{1}{2}\left(ln\right(1-sinx)-ln(1+sinx\left)\right)$

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}ln(1-sinx)-\frac{d}{dx}ln(1+sinx)$

$\frac{dy}{dx}=\frac{1}{2}(\frac{-cosx}{1-sinx}-\frac{cosx}{1+sinx})$

$\frac{dy}{dx}=\frac{1}{2}\left(\frac{-cosx(1+sinx)-cosx(1-sinx)}{(1-sinx)(1+sinx)}\right)$

$\frac{dy}{dx}=\frac{1}{2}\left(\frac{-cosx-cosxsinx-cosx+cossinx}{1-si{n}^{2}x}\right)$

$\frac{dy}{dx}=\frac{1}{2}\left(\frac{-2cosx}{co{s}^{2}x}\right)$

$\frac{dy}{dx}=\frac{-1}{cosx}$

$\frac{dy}{dx}=-secx$