Prove that- 1 3 -1 - 1 4 -1 -1 - -1

{(13)^ 1 (14)^ 1} ={ 3^1 4^1} ={3 4} = 1 . ...

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Prove that: ...

Question

Prove that:
${{(\frac{1}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}}^{- 1} = - 1$

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\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + . . . . . . . . . . . . . + \frac{1}{\sqrt{8} + \sqrt{9}}

2 {tan}^{- 1} \frac{1}{2} = {tan}^{- 1} \frac{4}{3}

n \geq 1

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n (n + 1)} = \frac{n}{n + 1}

\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{5}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}

= 2

{tan}^{- 1} \frac{1}{4} + {tan}^{- 1} \frac{2}{9} = \frac{1}{2} {cos}^{- 1} \frac{3}{5}

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