Question

The value of the sum $ \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k-1}}\cdot \sec {{2}^{k}} \right)} $ is
(a) $ \tan {{2}^{n}} $
(b) $ \tan {{2}^{n}}-1 $
(c) $ \tan {{2}^{n}}-\tan 1 $
(d) $ \cos {{2}^{n}}-\cos 2 $

Answer 1

Hint:First, we will convert the given equation i.e. $ \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k-1}}\cdot \sec {{2}^{k}} \right)} $ in sin and cos terms using the trigonometric formula $ \tan \theta =\dfrac{\sin \theta }{\cos \theta },\sec \theta =\dfrac{1}{\cos \theta } $ . Then, on further simplification we will have equation to be solved as $ \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k}}-\tan {{2}^{k-1}} \right)} $ . Thus, on substituting the values of k we will get our answer.

Complete step-by-step answer:
Here, we have to find the value of equation $ \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k-1}}\cdot \sec {{2}^{k}} \right)} $ . So, first we will convert tan and sec functions into the form of sin and cos terms using the formula $ \tan \theta =\dfrac{\sin \theta }{\cos \theta },\sec \theta =\dfrac{1}{\cos \theta } $ .
 So, on using the formula we can write it as
 $ \tan {{2}^{k-1}}\cdot \sec {{2}^{k}}=\dfrac{\sin {{2}^{k-1}}}{\cos {{2}^{k-1}}\cdot \cos {{2}^{k}}} $ …………………(1)
Now we can write $ {{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} $ . We know the rule that $ {{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}} $ . So, on solving the equation $ {{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} $ we get the value $ \dfrac{{{2}^{k}}}{{{2}^{1}}} $ as shown below.
 $ {{2}^{k}}-{{2}^{k-1}}={{2}^{k}}\left( 1-{{2}^{-1}} \right) $
On further solving, we get
 $ ={{2}^{k}}\left( 1-\dfrac{1}{2} \right)=\dfrac{{{2}^{k}}}{2} $
Thus, we got $ {{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} $ which is equal to $ \dfrac{{{2}^{k}}}{{{2}^{1}}} $ . So, we will put this value i.e. $ {{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} $ in equation (1) in numerator part only. So, we will get as
 $ =\dfrac{\sin \left( {{2}^{k}}-{{2}^{k-1}} \right)}{\cos {{2}^{k-1}}\cdot \cos {{2}^{k}}} $
Now, we will use the formula $ \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b $ . So, on using this we can write equation as
 $ =\dfrac{\sin {{2}^{k}}\cos {{2}^{k-1}}-\cos {{2}^{k}}\sin {{2}^{k-1}}}{\cos {{2}^{k-1}}\cos {{2}^{k}}} $
On further solving i.e. dividing each term with denominator we get
 $ =\dfrac{\sin {{2}^{k}}\cos {{2}^{k-1}}}{\cos {{2}^{k-1}}\cos {{2}^{k}}}-\dfrac{\cos {{2}^{k}}\sin {{2}^{k-1}}}{\cos {{2}^{k-1}}\cos {{2}^{k}}} $
Cancelling the same terms. We will get
 $ =\dfrac{\sin {{2}^{k}}}{\cos {{2}^{k}}}-\dfrac{\sin {{2}^{k-1}}}{\cos {{2}^{k-1}}} $
 $ =\tan {{2}^{k}}-\tan {{2}^{k-1}} $
Thus, we have our equation as $ \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k}}-\tan {{2}^{k-1}} \right)} $
On substituting the value of k as 1 in second term and k as n in first term, we will get
 $ =\tan {{2}^{n}}-\tan {{2}^{1-1}} $
 $ =\tan {{2}^{n}}-\tan {{2}^{0}} $
We know that any term raised to zero is equal to 1. So, here we will get answer as
 $ =\tan {{2}^{n}}-\tan 1 $
Thus, option (c) is the correct answer.

Note: Be careful while writing the value of $ {{2}^{k-1}} $ . If we directly write value as $ \dfrac{{{2}^{k}}}{{{2}^{1}}} $ and substituting in the equation, we will get equation as $ \dfrac{\sin \left( \dfrac{{{2}^{k}}}{2} \right)}{\cos \left( \dfrac{{{2}^{k}}}{2} \right)\cdot \cos {{2}^{k}}} $ and will not able to simply further by this. So, do not make this mistake. Otherwise will not be able to convert this equation into its original form of tan function.