Question

Find the sum of the given expression.
\[\tan x\tan 2x+\tan 2x\tan 3x+.......+\tan nx\tan \left( n+1 \right)x\].

Answer 1

Hint: Use the trigonometric formula of \[\tan \left( a-b \right)\]. Expand the formula and find the expression for \[\tan a\]and \[\tan b\]. Substitute in this expression for each term. And finally add and simplify them to get the sum.

Complete step-by-step answer:
We have been asked to find the sum of, \[\tan x\tan 2x+\tan 2x\tan 3x+.......+\tan nx\tan \left( n+1 \right)x\].
If we take the first expression, \[\tan x\tan 2x\], there is a difference of x in these 2 terms.
Similarly, taking the \[{{2}^{nd}}\]expression, \[\tan 2x\tan 3x\], there is a difference of x.
Now let us use the trigonometric formula to solve the expression.
\[\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}\]
Now let us rearrange the formula, by cross multiplying them.
\[\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}\]
\[\begin{align}
  & 1+\tan a\tan b=\dfrac{\tan a-\tan b}{\tan \left( a-b \right)} \\
 & \tan a\tan b=\dfrac{\tan a-\tan b}{\tan \left( a-b \right)}-1 \\
 & \therefore \tan a\tan b=\dfrac{\tan a-\tan b-\tan \left( a-b \right)}{\tan \left( a-b \right)}-(1) \\
\end{align}\]
Now according to this formula, if we are taking the value of \[\tan x\tan 2x\]in equation (1) then, taking \[\tan 2x\tan x\], where a = 2x, b = x.
\[\tan 2x\tan x=\dfrac{\tan 2x-\tan x-\tan \left( 2x-x \right)}{\tan \left( 2x-x \right)}\]
\[\begin{align}
  & =\dfrac{\tan 2x-\tan x-\tan x}{\tan x} \\
 & =\dfrac{\tan 2x-2\tan x}{\tan x} \\
\end{align}\]

Similarly, \[\tan 3x\tan x=\dfrac{\tan 3x-\tan 2x-\tan \left( 3x-2x \right)}{\tan \left( 3x-2x \right)}\]
\[=\dfrac{\tan 3x-\tan 2x-\tan x}{\tan x}\]
Similarly for, \[\tan \left( n+1 \right)x\tan x=\dfrac{\tan \left( n+1 \right)x-\tan nx-\tan \left( n+1-n \right)x}{\tan \left( n+1-n \right)x}\].
\[=\dfrac{\tan \left( n+1 \right)x-\tan nx-\tan x}{\tan x}\].
So, if we are adding all this expression together, \[\tan x\tan 2x+\tan 2x\tan 3x+.......+\tan nx\tan \left( n+1 \right)x\]
\[=\dfrac{\tan 2x-\tan x-\tan x+\tan 3x-\tan 2x-\tan x+\tan 4x-\tan 3x-\tan x+.......+\tan \left( n+1 \right)x-\tan x-\tan x}{\tan x}\]
By cancelling out the like terms, \[\tan 2x,\tan 3x\]etc
\[\begin{align}
  & =\dfrac{-\tan x-\tan x-\tan x-......+\tan \left( n+1 \right)x}{\tan x} \\
 & =\dfrac{-\left( n+1 \right)\tan x+\tan \left( n+1 \right)x}{\tan x} \\
 & =\dfrac{\tan \left( n+1 \right)x-\left( n+1 \right)\tan x}{\tan x} \\
\end{align}\]
So we got, \[\tan x\tan 2x+\tan 2x\tan 3x+.......+\tan nx\tan \left( n+1 \right)x\].
\[=\dfrac{\tan \left( n+1 \right)x-\left( n+1 \right)\tan x}{\tan x}\]

Note: In a question like this remember to use the trigonometric formula \[\tan \left( a-b \right)\]without which this expression can’t be solved. Learn the formulae of basic trigonometric identities so that you might get an idea of which formula to use for questions like these.