यदि `siny=xsin(a+y)` हो,तो सिद्ध कीजिए कि `(dy)/(dx)=(sin^(2)(a+y))/(sina)`

Explanation

`siny=xsin(a+y)`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`d/(dx)(siny)=d/(dx)[xsin(a+y)]`
`rArrcosy(dy)/(dx)=xd/(dx)sin(a+y)+sin(a+y)d/(dx)(x)`
`rArrcosy(dy)/(dx)=xcos(a+y)(dy)/(dx)+sin(a+y)`
`rArr{cosy-xcos(a+y)}(dy)/(dx)=sin(a+y)`
`rArr{cosy-(siny)/(sin(a+y)).cos(a+y)}(dy)/(dx)=sin(a+y)`
[समी. (1) से]
`rArr{(sin(a+y)cosy-sinycos(a+y))/(sin(a+y))}(dy)/(dx)=sin(a+y)`
`rArr(sin(a+y-y))/(sin(a+y))(dy)/(dx)=sin(a+y)`
`rArr(sina)/(sin(a+y))(dy)/(dx)=sin(a+y)`
`rArr(dy)/(dx)=(sin^(2)(a+y))/(sina)`
यही सिद्ध करना था|