The equation of the curve passing through the point 1,π4 and tangent at any point of which makes angle Tan-1yx-cos2yx with the X-axis is

Given θ= tan−1(yx−cos2(yx))⇒ tanθ=yx−cos2(yx)⇒ dydx=yx−cos2(yx)Put y= vxdydx=v+x dvdx⇒ v+x dvdx=v−cos2v⇒ x dvdx=−cos2v⇒ ∫sec2v dv=−∫1x dx⇒ tanv = −logx+c⇒ tan (yx)=−logx+cIt passes through (1, π4)⇒ tan(π4)=−log1+c⇒ c=1⇒ tan (yx)=−logx+1⇒ y= x tan−1(1− logx)

The equation of the curve passing through the point 1,π4 and tangent at any point of which makes angle Tan-1yx-cos2yx with the X-axis is

Given θ= tan−1(yx−cos2(yx))⇒ tanθ=yx−cos2(yx)⇒ dydx=yx−cos2(yx)Put y= vxdydx=v+x dvdx⇒ v+x dvdx=v−cos2v⇒ x dvdx=−cos2v⇒ ∫sec2v dv=−∫1x dx⇒ tanv = −logx+c⇒ tan (yx)=−logx+cIt passes through (1, π4)⇒ tan(π4)=−log1+c⇒ c=1⇒ tan (yx)=−logx+1⇒ y= x tan−1(1− logx)

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The equation of the curve passing through the point $(1, \frac{π}{4})$ and tangent at any point of which makes angle ${Tan}^{- 1} (\frac{y}{x} - \cos^{2} \frac{y}{x})$ with the X-axis is

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$y = {Tan}^{- 1} (\log \frac{e}{x})$

$y = x {Tan}^{- 1} (\log \frac{y}{x}) + 1$

$y = {xTan}^{- 1} (1 - logx)$

$y - x = {xTan}^{- 1} (1 - logx)$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n θ = \tan^{- 1} (\frac{y}{x} - \cos^{2} (\frac{y}{x})) \\ \Rightarrow \tan θ = \frac{y}{x} - \cos^{2} (\frac{y}{x}) \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} - \cos^{2} (\frac{y}{x}) \\ P u t y = v x \\ \frac{d y}{d x} = v + x \frac{d v}{d x} \\ \Rightarrow v + x \frac{d v}{d x} = v - \cos^{2} v \\ \Rightarrow x \frac{d v}{d x} = - \cos^{2} v \end{array}$

$\begin{array}{l} \Rightarrow \int \sec^{2} v d v = - \int \frac{1}{x} d x \\ \Rightarrow tanv = - l o g x + c \\ \Rightarrow \tan (\frac{y}{x}) = - \log x + c \\ I t p a s s e s t h r o u g h (1, \frac{π}{4}) \\ \Rightarrow \tan (\frac{π}{4}) = - \log 1 + c \\ \Rightarrow c = 1 \\ \Rightarrow \tan (\frac{y}{x}) = - \log x + 1 \\ \Rightarrow y = x \tan^{- 1} (1 - \log x) \end{array}$