Question

if y=log(under root 1-cos2x/1+cos2x) then show that dy/dx=2cosec2x

Answer 1

Y = Log $\frac{ \sqrt{1-cos2x} }{ \sqrt{1+cos2x} }$
   = Log $\frac{ \sqrt{2sin^{2}x } }{ \sqrt{2cos^{2}x }}$
   = Log $\frac{sinx}{cosx}$
   = log sinx - Log cosx
$\frac{dy}{dx}$ = $\frac{d}{dx}$ (log sinx - Log cosx)
                                   = $\frac{cosx}{sinx}$ + $\frac{sinx}{cosx}$
                                   = (sin²x + cos²x)/sinx.cosx
                                   = 2/2six.cosx
                                   = 2/sin2x
                                   = 2cosec2x

Answer 2

$\dfrac{dy}{dx}=2\csc 2x,$ proved.

Step-by-step explanation:

We have,

$y=\log \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}$

Show that, $\dfrac{dy}{dx}=2\csc 2x$

∴ $y=\log \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}$

⇒ $y=\log \sqrt{\dfrac{2\sin^2 2x}{2\cos^2 2x}}$

Using formula:

$1-\cos 2x=2\sin^2 2x$ and

$1+\cos 2x=2\cos^2 2x$

⇒ $y=\log \sqrt{\tan^2 2x}$

⇒ $y=\log ({\tan 2x})$

Differentiating both sides w.r.t. x, we get

$\dfrac{dy}{dx}=\dfrac{1}{\tan 2x} \dfrac{d(\tan 2x)}{dx}$

⇒ $\dfrac{dy}{dx}=\dfrac{1}{\tan 2x} 2\sec^ 2x$

⇒ $\dfrac{dy}{dx}=2\dfrac{\cos 2x}{\sin 2x}.\dfrac{1}{\cos^2 2x}$

⇒ $\dfrac{dy}{dx}=2\dfrac{1}{\sin 2x}$

⇒ $\dfrac{dy}{dx}=2\csc 2x,$ proved.