If #siny = xsin(a + y)#. Then how will you prove that #dy/dx = (sin^2(a + y))/sina#??

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Answer 1 · 2017-07-24T14:20:14

The Proof is given in the Explanation.

# siny=xsin(a+y).#

# :. x=siny/sin(a+y).#

Differentiating w.r.t. #y,# using the Quotient Rule, we have,

# dx/dy=[sin(a+y)*d/dy{siny}-siny*d/dx{sin(a+y)}]/sin^2(a+y),#

#={sin(a+y)cosy-sinycos(a+y)*d/dy(a+y)}/sin^2(a+y),..."[The Chain Rule],"#

#={sin(a+y)cosy-sinycos(a+y)}/sin^2(a+y),#

#=sin{(a+y)-y}/sin^2(a+y),#

#=sina/sin^2(a+y).#

# rArr dy/dx=1/{dx/dy}=sin^2(a+y)/sina.#

Enjoy Maths.!