If cos y = x cos (a + y) with cos a ≠ ± 1 then prove that

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Question:

If \cos y = x \cos(a + y), with \cos a \neq \pm 1, then prove that \dfrac{dy}{dx}=\dfrac{\cos^2(a + y)}{\sin a}. Hence show that \sin a \dfrac{d^2y}{dx^2}+ \sin 2(a + y) \dfrac{dy}{dx}=0. ……..      [CBSC  2016]

Solution:

Given, \cos y = x \cos(a + y)

\Rightarrow x = \dfrac{\cos y}{\cos (a + y)}

Differentiating with respect to y on both sides, we get

\dfrac{dx}{dy} = \dfrac{\cos (a + y)\times (-\sin y)-\cos y \times [-\sin(a + y)]}{\cos^2(a + y)}

\Rightarrow \dfrac{dx}{dy}= \dfrac{\cos y \sin (a + y) - \sin y \cos (a + y)}{\cos^2 y}

\Rightarrow \dfrac{dx}{dy} = \dfrac{\sin(a + y - y)}{\cos^2(a + y)}

\Rightarrow \dfrac{dx}{dy}=\dfrac{\sin a}{\cos^2(a + y)}

\Rightarrow \dfrac{dy}{dx} = \dfrac{\cos^2(a + y)}{\sin a}

Differentiating both sides w.r.t. x, we get

\dfrac{d^2 y}{dx^2} = \dfrac{1}{\sin a}\left[-2\cos (a + y).\sin (a + y).\dfrac{dy}{dx}\right]

\Rightarrow \sin a \dfrac{d^2}y{dx^2} = -\sin 2(a + y).\dfrac{dy}{dx}

\Rightarrow \sin a \dfrac{d^2y}{dx^2}+ \sin 2(a + y) \dfrac{dy}{dx}=0

Some other question:

Q 1: If \tan^{-1}(\frac{y}{x}) = \log \sqrt{x^2+y^2}, Prove that \dfrac{dy}{dx} = \dfrac{x + y}{x - y}.    ……..[CBSC 2020]

Solution: For solution click here

Q 2: If y^x = e^{y-x}, then prove that \dfrac{dy}{dx}=\dfrac{(1+\log y)^2}{\log y}  ……[CBSC 2013]

Solution: For solution click here

Q 3: Find the value of \dfrac{dy}{dx} at \theta = \dfrac{\pi}{4}, if x = ae^{\theta}(\sin \theta - \cos \theta) and y = ae^{\theta}(\sin \theta + \cos \theta).          ……..   [CBSC 2008, 2014]

Solution: For solution click here

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