$ y x = ( 2 + \sqrt { 3 } ) $ sind the value $ x ^ { 2 } + \frac { 1 } { x ^ { 2 } } $

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x^{2} + y^{2} = z^{2}

then find the value of

\frac{1}{{log}_{z - y} x} + \frac{1}{{log}_{z + y} x}

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Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

satisfy

y (2) = \frac{2}{\sqrt{3}}

STATEMENT-1:

y (x) = sec (s e c^{- 1} x - \frac{π}{6})

STATEMENT-2 :

y (x)

is given by

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

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Assertion :Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

satisfy

y (2) = 2 / \sqrt{3}

y (x) = sec ({sec}^{- 1} x - \frac{π}{6})

Reason:

y (x)

is given by

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

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Reason
$$y(x)$$ is given by $$\dfrac{1}{y}=\dfrac{2\sqrt 3}{x}-\sqrt{1-\dfrac{1}{x^2}}$$
Assertion
$$y(x)=\sec \left( \sec^{-1}x-\dfrac{\pi}{6}\right)$$ and
Let a solution $$y=y(x)$$ of the differential equation
$$x\sqrt{x^2-1}dy-y\sqrt{y^2-1}dx=0$$ satisfy $$y(2)=\dfrac{2}{\sqrt 3}$$

View Solution

Assertion :Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

satisfy

y (2) = \frac{2}{\sqrt{3}} .

y (x) = s e c ({sec}^{- 1} x - \frac{π}{6}),

Reason:

y (x)

is given by

\frac{1}{y} = \frac{2 \sqrt{3}}{c} - \sqrt{1 - \frac{1}{x^{2}}}

View Solution

\( y x = ( 2 + \sqrt { 3 } ) \) sind the value \( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \)