Question

The value of ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right) + {\tan ^{ - 1}}\left( {\dfrac{4}{{33}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{8}{{129}}} \right) + ..... + n$ terms is
A. ${\tan ^{ - 1}}{2^n} - \dfrac{\pi }{4}$
B. ${\tan ^{ - 1}}{2^n}$
C. ${\cot ^{ - 1}}{2^n}$
D. $\dfrac{{{{\sin }^{ - 1}}{2^n}}}{{{{\cos }^{ - 1}}{2^n}}}$

Answer 1

We write the terms of the given series by observing a pattern. Try to break each term like the value of formula ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y$. Apply this formula to each term and cancel all the terms having the same magnitude and opposite signs in the expansion. Write the values which are known to us using a table of trigonometric terms.

• ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$

Complete step-by-step answer:
We have to calculate the sum of the series ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right) + {\tan ^{ - 1}}\left( {\dfrac{4}{{33}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{8}{{129}}} \right) + ..... + n$terms
We can write the series as
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2 - 1}}{{1 + 2 \times 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{4 - 2}}{{1 + 4 \times 2}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{8 - 4}}{{1 + 8 \times 4}}} \right) + .... + {\tan ^{ - 1}}\left( {\dfrac{{{2^n} - {2^{n - 1}}}}{{1 + {2^n}{{.2}^{n - 1}}}}} \right)$
We try to write the second last term as well, so we are clear of what values sto cancel and what not to cancel.
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2 - 1}}{{1 + 2 \times 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{4 - 2}}{{1 + 4 \times 2}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{8 - 4}}{{1 + 8 \times 4}}} \right) + .... + {\tan ^{ - 1}}\left( {\dfrac{{{2^{n - 1}} - {2^{n - 2}}}}{{1 + {2^{n - 1}}{{.2}^{n - 2}}}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{{2^n} - {2^{n - 1}}}}{{1 + {2^n}{{.2}^{n - 1}}}}} \right)$
We know that each term matches the identity ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$. So series becomes
$\Rightarrow {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(4) - {\tan ^{ - 1}}(2) + {\tan ^{ - 1}}(8) - {\tan ^{ - 1}}(4) + .... + {\tan ^{ - 1}}({2^{n - 1}}) - {\tan ^{ - 1}}({2^{n - 2}}) + {\tan ^{ - 1}}({2^n}) - {\tan ^{ - 1}}({2^{n - 1}})$
Cancel all possible terms
$\Rightarrow - {\tan ^{ - 1}}(1) + {\tan ^{ - 1}}({2^n})$
We know that ${\tan ^{ - 1}}(1) = \dfrac{\pi }{4}$
$\Rightarrow {\tan ^{ - 1}}({2^n}) - \dfrac{\pi }{4}$
$\therefore$The sum of the series is ${\tan ^{ - 1}}({2^n}) - \dfrac{\pi }{4}$

$\therefore$Option A is correct.

Note:
Many students make the mistake of not cancelling the second last term after the expansion as they don’t see the negative term similar to that term so as to cancel the term. Keep in mind when we write the value of the second last term in the expansion, we get the negative term there and so we can cancel it. Students are advised to use the table of trigonometric terms if they don’t remember the values directly.
The table that tells us some basic values of trigonometric functions at common angles is given as

Angles (in degrees)	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3$	Not defined
cosec	Not defined	2	$\sqrt 2$	$\dfrac{2}{{\sqrt 3 }}$	1
sec	1	$\dfrac{2}{{\sqrt 3 }}$	$\sqrt 2$	2	Not defined
cot	Not defined	$\sqrt 3$	1	$\dfrac{1}{{\sqrt 3 }}$	0