If nbsp-secx-yx-y-a- nbsp-then nbsp-x nbsp-dy-y nbsp-dx-1-1

The correct option is C that, secx+yx y=a nbsp;⇒x+yx y=sec 1a1Applying nbsp;Componendo nbsp;and nbsp;Dividendo nbsp;both nbsp;sides,⇒2x2y=sec 1a+1se ...

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Standard XII

Mathematics

Implicit Differentiation

If nbsp;sec...

Question

If $sec (\frac{x + y}{x - y}) = a, then x dy - y dx =$

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Solution

The correct option is C 0.0
Given that,
$sec (\frac{x + y}{x - y}) = a \Rightarrow \frac{x + y}{x - y} = \frac{{sec}^{- 1} a}{1} Applying Componendo and Dividendo both sides, \Rightarrow \frac{2 x}{2 y} = \frac{{sec}^{- 1} a + 1}{{sec}^{- 1} a - 1} \Rightarrow \frac{y}{x} = \frac{{sec}^{- 1} a - 1}{{sec}^{- 1} a + 1} Differentiating both sides w . r . t . x, we get \Rightarrow \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^{2}} = 0 \Rightarrow x \frac{dy}{dx} - y = 0 \Rightarrow x dy - y dx = 0$

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If sec(x+yx−y)=a, then x dy−y dx=

The correct option is C 0.0Given that, sec(x+yx−y)=a ⇒x+yx−y=sec−1a1Applying Componendo and Dividendo both sides,⇒2x2y=sec−1a+1sec−1a−1⇒yx=sec−1a−1sec−1a+1Differentiating both sides w.r.t. x, we get⇒x⋅dydx−y⋅1x2=0⇒xdydx−y=0⇒x dy−y dx=0

If $sec (\frac{x + y}{x - y}) = a, then x dy - y dx =$