The angle between the curves y = \; x and y = x is

Correct Answer is : tan⁡−1(22)^{-1}(2 {2})tan−1(22)

Correct Answer is : tan⁡−1(22)^{-1}(2 {2})tan−1(22)

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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

Question : 120 of 160

Marks: +1, -0

y = \sin \; x

y = \cos x

Solution:

Given that,

y = \sin \; x

. . . (i)

and

\;\; y = \cos x

. . . (ii)

From Eqs. (i) and (ii),

\; \sin \; x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}

\;\; \text{ Slope } \; \; \;\; m_1 = \frac{dy}{dx} = \cos x,

\; m_2 = \frac{dy}{dx} = -\sin \; x

\therefore \;\; m_1 = \left( \frac{dy}{dx} \right)_{\frac{\pi}{4}} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}

\;\; \text{ and } \; \; \;\; m_2 = \left( \frac{dy}{dx} \right)_{\pi/4} = -\frac{1}{\sqrt{2}}

\therefore \;\; \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}

\Rightarrow \;\; \tan \theta = \frac{ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} }{ 1 - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} } = \frac{2}{\sqrt{2}} \cdot 2 = 2 \sqrt{2}

\therefore \;\; \theta = \tan^{-1}(2 \sqrt{2})

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