Question

❤️ Challenging Question ❤️

Differentiate :-

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Answer 1

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

We know that cos^2x + sin^2x = 1 and sin2x = 2cosxsinx

$= > \sqrt{\frac{cos^2x + sin^2x - 2cosxsinx}{cos^2x + sin^2x + 2cosxsinx}}$

We know that a^2 + b^2 - 2ab = (a - b)^2 and a^2 + b^2 + 2ab = (a + b)^2

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

$= > \frac{cosx - sinx}{cosx + sinx}$

Divide the numerator and denominator by 'cosx', we get

$= > \frac{cosx(1 - \frac{sinx}{cosx})}{cosx(1 + \frac{sinx}{cosx})}$

$= > \frac{1 - tanx}{1 + tanx}$

$= > \frac{1 - tanx}{1 + (1) * tanx}$

We know that tan(π/4) = 1.

$= > \frac{tan(\frac{\pi }{4} - tanx)}{1 + tan(\frac{\pi}{4}) * tan x}$

We know that tan(A - B) = tanA - tanB/(1 - tanA * tanB)

$= > tan(\frac{\pi}{4} - x)$

Differentiation:

$= > y = tan(\frac{\pi}{4} - x)$

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

Let v = (π/4 - x)

$= > \frac{d}{dv}(tan(v))\frac{d}{dx}(\frac{\pi}{4} - x)$

$= > sec^2(v)(-1)$

$= > sec^2(\frac{\pi}{4} - x)(-1)$

$= > -sec^2(\frac{\pi}{4} - x)$

So,

$= > \frac{dy}{dx} = -sec^2(\frac{\pi}{4} - x)$

(or)

$= > \boxed{\frac{dy}{dx} + sec^2(\frac{\pi}{4} - x) = 0}$

Hope it helps!