If `x=a{costheta+logtan""(theta)/(2)}" and "y=asintheta,` then `(dy)/(dx)` is equal to

Options

A. `cot theta`
B. `tantheta`
C. `sintheta`
D. `costheta`

Answer

B

Explanation

We have,
`x=a{costheta+logtan""(theta)/(2)}" and "y=asintheta`
`implies(dx)/(d theta)=a{-sintheta+("sec"^(2)(theta)/(2))/(2"tan"(theta)/(2))}" and "(dy)/(d theta)=acostheta`
`implies(dx)/(d theta)=a{-sintheta+(1)/(sintheta)}" and "(dy)/(d theta)=acostheta`
`implies" "(dx)/(d theta)=a(cos^(2)theta)/(sintheta)" and",(dy)/(d theta)=acostheta`
`(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(acostheta)/(a(cos^(2)theta)/(sintheta))=tantheta`