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Question: The value of the sum \( \sum\limits_{k=1}^{n}{\left( \tan {{2}^{k-1}}\cdot \sec {{2}^{k}} \right)} \...

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

Explanation

Solution

Hint:First, we will convert the given equation i.e. ∑k=1n(tan⁡2k−1⋅sec⁡2k)\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⁡θ=sin⁡θcos⁡θ,sec⁡θ=1cos⁡θ\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 ∑k=1n(tan⁡2k−tan⁡2k−1)\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 ∑k=1n(tan⁡2k−1⋅sec⁡2k)\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⁡θ=sin⁡θcos⁡θ,sec⁡θ=1cos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta },\sec \theta =\dfrac{1}{\cos \theta } .
So, on using the formula we can write it as
tan⁡2k−1⋅sec⁡2k=sin⁡2k−1cos⁡2k−1⋅cos⁡2k\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 2k−1=2k−2k−1{{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} . We know the rule that am−n=aman{{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}} . So, on solving the equation 2k−1=2k−2k−1{{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} we get the value 2k21\dfrac{{{2}^{k}}}{{{2}^{1}}} as shown below.
2k−2k−1=2k(1−2−1){{2}^{k}}-{{2}^{k-1}}={{2}^{k}}\left( 1-{{2}^{-1}} \right)
On further solving, we get
=2k(1−12)=2k2={{2}^{k}}\left( 1-\dfrac{1}{2} \right)=\dfrac{{{2}^{k}}}{2}
Thus, we got 2k−1=2k−2k−1{{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} which is equal to 2k21\dfrac{{{2}^{k}}}{{{2}^{1}}} . So, we will put this value i.e. 2k−1=2k−2k−1{{2}^{k-1}}={{2}^{k}}-{{2}^{k-1}} in equation (1) in numerator part only. So, we will get as
=sin⁡(2k−2k−1)cos⁡2k−1⋅cos⁡2k=\dfrac{\sin \left( {{2}^{k}}-{{2}^{k-1}} \right)}{\cos {{2}^{k-1}}\cdot \cos {{2}^{k}}}
Now, we will use the formula sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b . So, on using this we can write equation as
=sin⁡2kcos⁡2k−1−cos⁡2ksin⁡2k−1cos⁡2k−1cos⁡2k=\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
=sin⁡2kcos⁡2k−1cos⁡2k−1cos⁡2k−cos⁡2ksin⁡2k−1cos⁡2k−1cos⁡2k=\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
=sin⁡2kcos⁡2k−sin⁡2k−1cos⁡2k−1=\dfrac{\sin {{2}^{k}}}{\cos {{2}^{k}}}-\dfrac{\sin {{2}^{k-1}}}{\cos {{2}^{k-1}}}
=tan⁡2k−tan⁡2k−1=\tan {{2}^{k}}-\tan {{2}^{k-1}}
Thus, we have our equation as ∑k=1n(tan⁡2k−tan⁡2k−1)\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⁡2n−tan⁡21−1=\tan {{2}^{n}}-\tan {{2}^{1-1}}
=tan⁡2n−tan⁡20=\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⁡2n−tan⁡1=\tan {{2}^{n}}-\tan 1
Thus, option (c) is the correct answer.

Note: Be careful while writing the value of 2k−1{{2}^{k-1}} . If we directly write value as 2k21\dfrac{{{2}^{k}}}{{{2}^{1}}} and substituting in the equation, we will get equation as sin⁡(2k2)cos⁡(2k2)⋅cos⁡2k\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.