Question

x= a ( cost +logtan t/2 y = asin t find dy / dx​

Answer 1

Answer: $\dfrac{dy}{dx} = \tan t$

Explanation:

$x = a(\cos t + \log \tan\dfrac{t}2)$

Differentiate with respect to t,

$\dfrac{dx}{dt} =a \dfrac{d{(\cos t + \log\tan \dfrac{t}2)}}{dt}$  

$= a[\dfrac{d(\cos t)}{dt} + \dfrac{d(\log\tan \dfrac{t}2)}{dt}]$

$= a[ -\sin t + \dfrac{1}{\tan  \dfrac{t}2 }\times sec^2\dfrac{t}2\times \dfrac12]$

$= a [ -\sin t + \dfrac12 \dfrac{1}{\sin \dfrac{t}2\cdot\cos \dfrac{t}2 }]$

Since $2\sin A\cdot\cos A = \sin2 A$

$= a [ -\sin t + \dfrac1{\sin t}]$

$= a \dfrac{ 1 - \sin^2 t}{\sin t}$

$= a\dfrac{\cos^2t}{\sin t}$

$= a\cot t\cdot \cos t .......(1)$

again, $y = a\sin t$

$\dfrac{dy}{dt} = a\cos t .......(2)$

now, $\dfrac{dy}{dx} =\dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$ [from (1) and (2)]  

$=\dfrac {a\cos t}{a\cot t.\cos t}$  

$= \dfrac{1}{\cot t }$

Hence, $\dfrac{dy}{dx} = \tan t$