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Hint: - Use \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\]

Given:

$\tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ......n{\text{ terms}}$

Last term of this series is \[\tan nx\tan \left( {\left( {n + 1} \right)x} \right)\]

\[ \Rightarrow \tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ....... + \tan nx\tan \left( {\left( {n + 1} \right)x} \right)\]

Now, as we know

\[

\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}} \\

\Rightarrow 1 + \tan a\tan b = \dfrac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} \\

\Rightarrow \tan a\tan b = \dfrac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} - 1 \\

\Rightarrow \tan a\tan b = \dfrac{{\tan a - \tan b - \tan \left( {a - b} \right)}}{{\tan \left( {a - b} \right)}} \\

\]

So, applying this property in each term of the given series we get

\[

\Rightarrow \tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ....... + \tan nx\tan \left( {\left( {n + 1} \right)x} \right) \\

\Rightarrow \left( {\dfrac{{\tan 2x - \tan x - \tan \left( {2x - x} \right)}}{{\tan \left( {2x - x} \right)}}} \right) + \left( {\dfrac{{\tan 3x - \tan 2x - \tan \left( {3x - 2x} \right)}}{{\tan \left( {3x - 2x} \right)}}} \right) + \left( {\dfrac{{\tan 4x - \tan 3x - \tan \left( {4x - 3x} \right)}}{{\tan \left( {4x - 3x} \right)}}} \right) + ........... \\

.............. + \left( {\dfrac{{\tan \left( {\left( {n + 1} \right)x} \right) - \tan nx - \tan \left( {nx + x - nx} \right)}}{{\tan \left( {nx + x - nx} \right)}}} \right) \\

\Rightarrow \dfrac{{\tan 2x - \tan x - \tan x + \tan 3x - \tan 2x - \tan x + \tan 4x - \tan 3x - \tan x + .................. + \tan \left( {\left( {n + 1} \right)x} \right) - \tan nx - \tan x}}{{\tan x}} \\

\]

Now, we see many terms are cancel out the remaining terms are,

\[

\Rightarrow \dfrac{{ - \tan x + \left( { - \tan x - \tan x - ..................... + \tan \left( {\left( {n + 1} \right)x} \right) - \tan x} \right)}}{{\tan x}} \\

\Rightarrow \dfrac{{ - \tan x + \left( { - \tan x - \tan x - ..................... - \tan x} \right) + \tan \left( {\left( {n + 1} \right)x} \right)}}{{\tan x}} \\

\Rightarrow \dfrac{{ - \tan x - \tan x\left( {1 + 1 + 1 + ....................... + 1} \right) + \tan \left( {\left( {n + 1} \right)x} \right)}}{{\tan x}} \\

\]

As we know sum of 1 up to n terms is n

\[ \Rightarrow \dfrac{{\left( {\tan \left( {\left( {n + 1} \right)x} \right) - n\tan x} \right) - \tan x}}{{\tan x}} = \dfrac{{\tan \left( {\left( {n + 1} \right)x} \right) - \left( {n + 1} \right)\tan x}}{{\tan x}}\]

So, this is the required sum of the given series.

Note: - In such types of questions always remember to always remember all the basic formulas of trigonometric identities, then apply this formula in the given equation and simplify, we will get the required answer.

Given:

$\tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ......n{\text{ terms}}$

Last term of this series is \[\tan nx\tan \left( {\left( {n + 1} \right)x} \right)\]

\[ \Rightarrow \tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ....... + \tan nx\tan \left( {\left( {n + 1} \right)x} \right)\]

Now, as we know

\[

\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}} \\

\Rightarrow 1 + \tan a\tan b = \dfrac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} \\

\Rightarrow \tan a\tan b = \dfrac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} - 1 \\

\Rightarrow \tan a\tan b = \dfrac{{\tan a - \tan b - \tan \left( {a - b} \right)}}{{\tan \left( {a - b} \right)}} \\

\]

So, applying this property in each term of the given series we get

\[

\Rightarrow \tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ....... + \tan nx\tan \left( {\left( {n + 1} \right)x} \right) \\

\Rightarrow \left( {\dfrac{{\tan 2x - \tan x - \tan \left( {2x - x} \right)}}{{\tan \left( {2x - x} \right)}}} \right) + \left( {\dfrac{{\tan 3x - \tan 2x - \tan \left( {3x - 2x} \right)}}{{\tan \left( {3x - 2x} \right)}}} \right) + \left( {\dfrac{{\tan 4x - \tan 3x - \tan \left( {4x - 3x} \right)}}{{\tan \left( {4x - 3x} \right)}}} \right) + ........... \\

.............. + \left( {\dfrac{{\tan \left( {\left( {n + 1} \right)x} \right) - \tan nx - \tan \left( {nx + x - nx} \right)}}{{\tan \left( {nx + x - nx} \right)}}} \right) \\

\Rightarrow \dfrac{{\tan 2x - \tan x - \tan x + \tan 3x - \tan 2x - \tan x + \tan 4x - \tan 3x - \tan x + .................. + \tan \left( {\left( {n + 1} \right)x} \right) - \tan nx - \tan x}}{{\tan x}} \\

\]

Now, we see many terms are cancel out the remaining terms are,

\[

\Rightarrow \dfrac{{ - \tan x + \left( { - \tan x - \tan x - ..................... + \tan \left( {\left( {n + 1} \right)x} \right) - \tan x} \right)}}{{\tan x}} \\

\Rightarrow \dfrac{{ - \tan x + \left( { - \tan x - \tan x - ..................... - \tan x} \right) + \tan \left( {\left( {n + 1} \right)x} \right)}}{{\tan x}} \\

\Rightarrow \dfrac{{ - \tan x - \tan x\left( {1 + 1 + 1 + ....................... + 1} \right) + \tan \left( {\left( {n + 1} \right)x} \right)}}{{\tan x}} \\

\]

As we know sum of 1 up to n terms is n

\[ \Rightarrow \dfrac{{\left( {\tan \left( {\left( {n + 1} \right)x} \right) - n\tan x} \right) - \tan x}}{{\tan x}} = \dfrac{{\tan \left( {\left( {n + 1} \right)x} \right) - \left( {n + 1} \right)\tan x}}{{\tan x}}\]

So, this is the required sum of the given series.

Note: - In such types of questions always remember to always remember all the basic formulas of trigonometric identities, then apply this formula in the given equation and simplify, we will get the required answer.

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