Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

We use the identity $\tan \theta =\cot \theta -2\cot (2\theta )$. Multiplying by $2^k$, we get $2^k\tan (2^k\theta )=2^k\cot (2^k\theta )-2^{k+1}\cot (2^{k+1}\theta )$. The sum $S_n$ becomes a telescoping series:$$S_n=\sum _{k=0}^{n-1}(2^k\cot (2^k{x}_n)-2^{k+1}\cot (2^{k+1}{x}_n))=2^0\cot (2^0{x}_n)-2^n\cot (2^n{x}_n)=\cot (x_n)-2^n\cot (2^n{x}_n)$$ Given$x_n=\frac{\pi }{2^n+1}$. Then $2^n{x}_n=\frac{2^n\pi }{2^n+1}=\frac{(2^n+1-1)\pi }{2^n+1}=\pi -\frac{\pi }{2^n+1}=\pi -x_n$. Therefore, $\cot (2^n{x}_n)=\cot (\pi -x_n)=-\cot (x_n)$. Substituting this, $S_n=\cot (x_n)-2^n(-\cot (x_n))=(1+2^n)\cot (x_n)$. Thus, option A is correct. For$n=2$, $x_2=\pi /(2^2+1)=\pi /5$. So $S_2=(1+2^2)\cot (\pi /5)=5\cot (\pi /5)$. Option C is correct. For$n=3$, $x_3=\pi /(2^3+1)=\pi /9$. So $S_3=(1+2^3)\cot (\pi /9)=9\cot (\pi /9)$. Option D is correct.