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Expansion of (x^2 - y^2)

If x-1 x=-1, ...

Question

If $x - \frac{1}{x} = - 1$ , find the value of $x^{2} + \frac{1}{x^{2}}$

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Solution

In the given problem, we have to find

Given

On squaring both sides we get,

We shall use the identity

Hence the value of is.

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Similar questions

x = 3 - 2 \sqrt{2}

, find the value of:
i)

\frac{1}{x}

x + \frac{1}{x}

x - \frac{1}{x}

\sqrt{x} + \frac{1}{\sqrt{x}}

\sqrt{x} - \frac{1}{\sqrt{x}}

x^{2} + \frac{1}{x^{2}}

x^{2} - \frac{1}{x^{2}}

x + \frac{1}{x} = 11,

find the value of

x^{2} + \frac{1}{x^{2}}

x + \frac{1}{x} = 20,

find the value of

x^{2} + \frac{1}{x^{2}} .

x^{2} + \frac{1}{x^{2}} = 18,

find the values of

x + \frac{1}{x} and x - \frac{1}{x} .

(x - 1) (x^{2} + 1) > 0

, then find the value of

sin (\frac{1}{2} {tan}^{- 1} \frac{2 x}{1 - x^{2}} - {tan}^{- 1} x)

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