Question

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \left(P\right)& \frac{{\prod }_{k=1}^{100}(1-ta{n}^2\frac{2^k\pi }{2^{100}+1})}{2^{100}}\text{is}& 1.& 1\\ \left(Q\right)& {\prod }_{k=1}^{100}(1+2cos\frac{2\pi 3^k}{3^{100}+1})\text{is}& 2.& -1\\ \left(R\right)& \text{Let n be a positive integer and}& 3.2& \\ & \text{Let x be a real number different}& & \\ & \text{from}{2}^{k+1},k=1,\dots \dots n,\text{then}& & \\ & \left[(1+2cos\frac{x}{2^{100}}){\prod }_{k=1}^{100}(1-2cos\frac{x}{2^k})\right]& & \\ & -2cosx\text{is equal to}& & \\ \left(S\right)& \text{The value of}{\sum }_{n=1}^{\infty }\frac{1}{2^n}tan\left(\frac{2}{2^n}\right)+cot2is& 4.& 3\\ & & 5.& \frac{1}{2}\end{array}$

• A. $P\to 2Q\to 1R\to 2S\to 4$
• B. $P\to 2Q\to 1R\to 1S\to 5$
• C. $P\to 4Q\to 2R\to 1S\to 3$
• D. $P\to 4Q\to 2R\to 1S\to 2$

Answer 1

The correct option is B $P\to 2Q\to 1R\to 1S\to 5$

(P) $1-ta{n}^{2}x=\frac{2tanx}{tan2x}\text{So}(1-ta{n}^2\frac{2^K\pi }{2^{100}+1})=\frac{2tan\frac{2^k\pi }{2^{100}+1}}{tan\left(\frac{2^{k+1}\pi }{2^{100}+1}\right)}$
On multiplying
$\frac{\frac{2tan\frac{2\pi }{2^{100}+1}}{tan\left(\frac{2^2\pi }{2^{100}+1}\right)}\times \frac{2tan\frac{2^2\pi }{2^{100}+1}}{tan\left(\frac{2^3\pi }{2^{100}+1}\right)}\times \dots \dots \times \frac{2tan\frac{2^{100}\pi }{2^{100}+1}}{tan\left(\frac{2^{101}\pi }{2^{100}+1}\right)}}{2^{100}}$
$=\frac{\text{/}{2}^{100}tan\left(\frac{2\pi }{2^{100}+1}\right)}{\text{/}{2}^{100}tan\left(\frac{2^{101}\pi }{2^{100}+1}\right)}=\frac{tan(2\pi -\frac{2^{101}\pi }{2^{100}+1})}{tan\left(\frac{2^{101}\pi }{2^{100}+1}\right)}=-1$

(Q) $1+2cos2a_k=1+2(1-2si{n}^2{a}_k)$
$=3-4si{n}^2{a}_k$
$=\frac{sin3a_k}{sin{a}_k}$, where $a_k=\frac{3^{k}x}{3^{n}H}$
So ${\prod }^1{\infty }_{k=1}(1+2cos\frac{2\pi 3^k}{3^{100}H})=\frac{sin3a_1}{sin{a}_1}\times \frac{sin3a_2}{sin{a}_2}\times \dots \dots \times \frac{sin3a_{100}}{sin{a}_{100}}$
$=\frac{sin\left(\frac{\text{/}{3}^2\pi }{\text{/}{3}^{100}H}\right)}{sin\left(\frac{3\pi }{3^{100}H}\right)}\times \frac{sin\left(\frac{3^3\pi }{3^{100}H}\right)}{sin\left(\frac{\text{/}{3}^2\pi }{\text{/}{3}^{100}H}\right)}\times \dots \dots \times \frac{sin\left(\frac{3^{101\pi }}{3^{100}H}\right)}{sin\left(\frac{3^{100\pi }}{3^{100}H}\right)}$
$=\frac{sin\left(\frac{3^{101\pi }}{3^{100}H}\right)}{sin\left(\frac{3\pi }{3^{100}H}\right)}=1$

(R) $1-2cosa=\frac{1-4co{s}^{2}a}{1+2cosa}=\frac{1-2(1+cos2a)}{1+2cosa}=-\left(\frac{1+2cos2a}{1+2cosa}\right)$
So ${\prod }_{k=1}^{100}(1-2cos\frac{x}{2^k})=-\left(\frac{1+2cosx}{1+2cos\frac{x}{2}}\right)\times -\left(\frac{1+2cos\frac{x}{2}}{1+2cos\frac{x}{4}}\right)\times \dots \dots \frac{-(1+2cos\frac{x}{2^{99}})}{(1+2cos\frac{x}{2^{100}})}$
$=\frac{1+2cosx}{1+2cos\frac{x}{2^{100}}}$
So $\left[(1+2cos\frac{x}{2^{100}}){\prod }_{k=1}^{100}(1-2cos\frac{x}{2^k})\right]-2cosx=1$

(S) $tanx=cotx-2cot2x$
${\sum }_{n=1}^{\infty }\frac{1}{2^n}tan\left(\frac{2}{2^n}\right)=\frac{1}{2}tan\left(\frac{2}{2^1}\right)+\frac{1}{2^2}tan\left(\frac{2}{2^2}\right)+\frac{1}{2^3}tan\left(\frac{2}{2^3}\right)+\dots \dots$
$=\frac{1}{2}[cot\text{/}1-2cot2]+\frac{1}{2^2}[cot\left(\frac{1}{2}\right)-\text{/}2cot\left(1\right)]+\frac{1}{2^3}[cot\frac{1}{2^2}-2cot\left(\frac{1}{2}\right)]+\dots \dots$
$=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}cot\left(\frac{2}{2^n}\right)-cot2$
$=\frac{1}{2}-cot2$
So, ${\sum }_{n=1}^{\infty }\frac{1}{2^n}tan\left(\frac{x}{2^n}\right)+cot2=\frac{1}{2}$