# How to prove the following? #cot^(-1)7+cot^(-1)8+cot^(-1)18=cot^(-1)3#

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To prove the identity ( \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18) = \cot^{-1}(3) ), we can use the tangent addition formula, which states that for any two angles ( \alpha ) and ( \beta ):

[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} ]

First, let's rewrite each term in terms of tangent:

[ \cot^{-1}(7) = \tan^{-1}\left(\frac{1}{7}\right) ] [ \cot^{-1}(8) = \tan^{-1}\left(\frac{1}{8}\right) ] [ \cot^{-1}(18) = \tan^{-1}\left(\frac{1}{18}\right) ]

Now, apply the tangent addition formula repeatedly:

[ \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}}\right) ]

[ = \tan^{-1}\left(\frac{15}{56}\right) ]

[ \tan^{-1}\left(\frac{15}{56}\right) + \tan^{-1}\left(\frac{1}{18}\right) = \tan^{-1}\left(\frac{\frac{15}{56} + \frac{1}{18}}{1 - \frac{15}{56} \cdot \frac{1}{18}}\right) ]

[ = \tan^{-1}(3) ]

Thus, ( \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18) = \cot^{-1}(3) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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