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

& 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{aligned}$$ Now according to this formula, if we are taking the value of $$\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{aligned} & =\dfrac{\tan 2x-\tan x-\tan x}{\tan x} \\\ & =\dfrac{\tan 2x-2\tan x}{\tan x} \\\ \end{aligned}$$ Similarly, $$\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{aligned} & =\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{aligned}$$ So we got, $$\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.