यदि ` sin y = x cos ( a+y) , ` तो सिद्ध कीजिए , `(dy)/(dx) = (cos^(2) (a+y))/(cos a) ` तथा ` x = 0 " " (dy)/(dx) = cos a ` .

Explanation

` sin y = x cos (a+y) `
` rArr x = (sin y) /( cos ( a+ y) )`
` rArr x = (sin y) /( cos ( a+ y) )`
` 1 = cos (a+ y) cos y (dy)/(dx) - sin y *{-sin ( a+y) } (dy)/(dx)`
` 1 = ({ cos ( a+y) * cos Y + sin (a+y) *sin } (dy)/(dx))/(cos^(2) (a+y))`
` 1 + (cos [ a+y -y](dy)/(dx))/(cos^(2) (a+y) )`
` 1 + (cos [ a+y -y](dy)/(dx))/(cos^(2) (a+y) )`
` rArr (dy)/(dx) = cos^(2) (a+y)/(cosa)`
पुनः जब ` x=0`
तब ` sin y = 0 rArr y = n pi `
` x = 0 " " (dy)/(dx) = cos^(2) (a+ npi)/(cosa)`
`rArr (dy)/(dx) = (cos^(2) a)/(cosa) = cos a `