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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

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Question : 65 of 160

Marks: +1, -0

If

x=\cos 2 t+\log (\tan t)

y=2 t+\cot 2 t

\;\frac{dy}{dx}=

[TG EAPCET 9 May 2024 Shift 2]

$\;\tan 2t$
$\;-\csc 2t$
$\;-\cot 2t$
$\;\sec 2t$

Solution:

Given,

x=\cos 2 t+\log (\tan t)

and

\;\; y=2 t+\cot 2 t

\;\therefore\;\;\;\frac{dx}{dt}=(-\sin 2t)(2)

\;+\;\frac{1}{\tan t}\sec^2 t=-2\sin 2t+\;\frac{1}{\sin t\cos t}

\;\;\text{ and }\;\;\frac{dy}{dx}=2-(\csc^2 2t(2).

\;=2(1-\csc^2 2t)

\;\therefore\;\;\;\frac{dy}{dx}=\;\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\;\frac{2(1-\csc^2 2t)}{-2\sin 2t+\frac{1}{\sin t\cos t}}

\;=\;\frac{2(-\cot^2 2t)}{-2\sin 2t+\frac{2}{\sin 2t}}

\;=\;\frac{\frac{\cos^2 2t}{\sin^2 2t}}{\sin 2t-\frac{1}{\sin 2t}}

\;=\;\frac{\cos^2 2t}{(\sin^2 2t-1)(\sin 2t)}

\;=\;\frac{\cos^2 2t}{-\cos^2 2t\sin 2t}=-\csc 2t

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