Question

Y is equal to 10 inverse under root 1 - 2 Sin 2x upon 1 + sin 2x find dy/dx​

Answer 1

Correct question:

$y = { \tan }^{ - 1} \sqrt{ \dfrac{1 - \sin2x }{1 + \sin2x } }$

Find $\dfrac{dy}{dx}$

Answer:

$\large\boxed{\sf{-1}}$

Step-by-step explanation:

Given that,

$y = { \tan }^{ - 1} \sqrt{ \dfrac{1 - \sin2x }{1 + \sin2x } }$

But, we know that,

$\sin(2x) = \dfrac{2 \tan(x) }{1 + { \tan }^{2} x}$

Substituting this value and simplifying, we get,

$= > y = { \tan }^{ - 1} \sqrt{ \dfrac{1 - \frac{2 \tan(x) }{1 + { \tan }^{2}x } }{1 + \frac{2 \tan(x) }{1 + { \tan }^{2} x} } } \\ \\ = > y = { \tan }^{ - 1} \sqrt{ \dfrac{1 + { \tan }^{2} x - 2 \tan(x) }{1 + { \tan }^{2} x + 2 \tan(x) } } \\ \\ = > y = { \tan}^{ - 1} \sqrt{ \dfrac{ {(1 - \tan x) }^{2} }{ {(1 + \tan x) }^{2} } } \\ \\ = > y = { \tan}^{ - 1} \sqrt{ {( \frac{1 - \tan x}{1 + \tan x } )}^{2} } \\ \\ = > y = { \tan}^{ - 1} ( \frac{1 - \tan x }{1 + \tan x} ) \\ \\ = > y = { \tan }^{ - 1} ( \frac{ \tan \frac{\pi}{4} - \tan x }{1 + \tan \frac{\pi}{4} \tan x } ) \\ \\ = > y = { \tan }^{ - 1}( {\tan( \frac{\pi}{4} - x ) } ) \\ \\ = > y = \frac{\pi}{4} - x$

Now, differentiating both sides wrt x, we get,

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} ( \frac{\pi}{4} - x) \\ \\ = > \frac{dy}{dx} = - 1$

Hence, the required value is -1.