If x = cos 2t + log(tan t) and y = 2t + cot 2 | vedclass

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(B) Given $x = \cos 2t + \log(	an t)$ and $y = 2t + \cot 2t$. First, differentiate $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2t + \cot 2

Vedclass · Accepted Answer

$-\operatorname{cosec} 2t$

Vedclass · Answer

(B) Given $x = \cos 2t + \log(	an t)$ and $y = 2t + \cot 2t$. First, differentiate $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2t + \cot 2t) = 2 - 2\operatorname{cosec}^2 2t = -2(\operatorname{cosec}^2 2t - 1) = -2\cot^2 2t$. Next, differentiate $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(\cos 2t + \log(	an t)) = -2\sin 2t + \frac{1}{	an t} \cdot \sec^2 t = -2\sin 2t + \frac{\cos t}{\sin t} \cdot \frac{1}{\cos^2 t} = -2\sin 2t + \frac{1}{\sin t \cos t}$. Since $\sin 2t = 2\sin t \cos t$, we have $\frac{1}{\sin t \cos t} = \frac{2}{\sin 2t} = 2\operatorname{cosec} 2t$. So, $\frac{dx}{dt} = -2\sin 2t + 2\operatorname{cosec} 2t = 2(\operatorname{cosec} 2t - \sin 2t) = 2\left(\frac{1 - \sin^2 2t}{\sin 2t}ight) = 2\frac{\cos^2 2t}{\sin 2t} = 2\cot 2t \cos 2t$. Finally, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\cot^2 2t}{2\cot 2t \cos 2t} = -\frac{\cot 2t}{\cos 2t} = -\frac{\cos 2t}{\sin 2t} \cdot \frac{1}{\cos 2t} = -\frac{1}{\sin 2t} = -\operatorname{cosec} 2t$.

If $x = \cos 2t + \log(\tan t)$ and $y = 2t + \cot 2t$, then $\frac{dy}{dx} = $

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