Solve:
$\tan x\tan 2x + \tan 2x\tan 3x + \tan 3x\tan 4x + ......n{\text{ terms}} = $
Last updated date: 20th Mar 2023
•
Total views: 308.1k
•
Views today: 5.88k
Answer
308.1k+ views
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.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
