Answer: [tex] \dfrac{dy}{dx} = \tan t [/tex]
Explanation:
[tex]x = a(\cos t + \log \tan\dfrac{t}2) [/tex]
Differentiate with respect to t,
[tex] \dfrac{dx}{dt} =a \dfrac{d{(\cos t + \log\tan \dfrac{t}2)}}{dt}[/tex]
[tex]= a[\dfrac{d(\cos t)}{dt} + \dfrac{d(\log\tan \dfrac{t}2)}{dt}][/tex]
[tex]= a[ -\sin t + \dfrac{1}{\tan \dfrac{t}2 }\times sec^2\dfrac{t}2\times \dfrac12] [/tex]
[tex]= a [ -\sin t + \dfrac12 \dfrac{1}{\sin \dfrac{t}2\cdot\cos \dfrac{t}2 }] [/tex]
Since [tex] 2\sin A\cdot\cos A = \sin2 A [/tex]
[tex]= a [ -\sin t + \dfrac1{\sin t}][/tex]
[tex]= a \dfrac{ 1 - \sin^2 t}{\sin t}[/tex]
[tex]= a\dfrac{\cos^2t}{\sin t}[/tex]
[tex]= a\cot t\cdot \cos t .......(1) [/tex]
again, [tex]y = a\sin t[/tex]
[tex]\dfrac{dy}{dt} = a\cos t .......(2)[/tex]
now, [tex] \dfrac{dy}{dx} =\dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}[/tex] [from (1) and (2)]
[tex]=\dfrac {a\cos t}{a\cot t.\cos t}[/tex]
[tex]= \dfrac{1}{\cot t } [/tex]
Hence, [tex] \dfrac{dy}{dx} = \tan t [/tex]
