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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 66 of 160

Marks: +1, -0

If

x=a[\cos \theta+\log (\tan (\theta / 2))]

y=a \sin \theta

\frac{dy}{dx}=

______

$\cot \theta$
$\tan \theta$
$\sin \theta$
$\cos \theta$

Solution:

x=a[\cos \theta+\log (\tan \theta/ 2)]

and

y=a \sin \theta

\therefore \frac{dx}{d\theta}=a\left[-\sin \theta+\frac{1}{\tan (\theta/ 2)} \cdot \sec ^{2} \theta/ 2 \cdot \frac{1}{2}\right]

[\because \frac{d}{dx} \tan x=\sec ^{2} x

and

\frac{d}{dx} \cos x=-\sin x

and

\frac{d}{dx} \log x=\frac{1}{x}]

=a \left[-\sin \theta+\frac{1}{2 \sin \theta/ 2 \cdot \cos \theta/ 2}\right]

=a\left[-\sin \theta+\frac{1}{\sin \theta}\right]

\frac{dx}{d\theta}=a\left[\frac{1-\sin ^{2} \theta}{\sin \theta}\right]=\frac{a \cos ^{2} \theta}{\sin \theta}

...(i)

and

y =a \sin \theta

\Rightarrow \frac{dy}{d\theta}=a \cos \theta

...(ii)

Eq. (ii) divide by Eq. (i)

\Rightarrow \frac{dy}{dx}=\frac{a \cos \theta}{\frac{a \cos ^{2} \theta}{\sin \theta}}

=\frac{\sin \theta}{\cos \theta}=\tan \theta

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