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Question: If x sin (α + y) = sin y and y′ = \(\frac{m}{(x^{2} + 2nx + 1)}\), then...

If x sin (α + y) = sin y and y′ = m(x2+2nx+1)\frac{m}{(x^{2} + 2nx + 1)}, then

A

m – n = 1

B

m + n = 1

C

m2 + n2 = 1

D

m = n

Answer

m2 + n2 = 1

Explanation

Solution

tan y = xsin⁡α1−xcos⁡α\frac{x\sin\alpha}{1 - x\cos\alpha}

On differentiating wrt x, we get

dydx\frac{dy}{dx}= sin⁡α(1−xcos⁡α)\frac{\sin\alpha}{(1 - x\cos\alpha)}

⇒ m = sin α, n = – cos α

∴ m2 + n2 =1