If x - 1x - 3- then x2 - 1x2 equals -

The correct option is D 7 Given, : #160; ( x + 1x ) = 3 Squaring both sides, we get, ( x + 1x )^2 = 9 ⇒ x^2 + #160;1x^2 + 2× x×1x = 9 ...

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(a + b)^2 Expansion and Visualisation

If x + 1x =...

Question

If $(x + \frac{1}{x}) = 3$ , then $(x^{2} + \frac{1}{x^{2}})$ equals _______.

\frac{10}{3}

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\frac{82}{9}

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7

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11

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y = {tan}^{- 1} \frac{1}{1 + x + x^{2}} + {tan}^{- 1} \frac{1}{x^{2} + 3 x + 3} + {tan}^{- 1} \frac{1}{x^{2} + 5 x + 7} + . . . +

y^{'} (0)

y = t a n^{- 1} \frac{1}{1 + x + x^{2}} + t a n^{- 1} \frac{1}{x^{2} + 3 x + 3} + t a n^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots +

y^{'} (0)

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

+ {tan}^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots +

n

3 s i n^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 t a n^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

| x | < 1

x

{cot}^{- 1} (\frac{1 - x^{2}}{2 x}) + {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = \frac{2 π}{3}, x > 0

x \neq 1

x =

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