Question

Using the principle of mathematical induction, find $tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +....$ to $n$ terms:

• A. $tan\alpha -2^n\cdot tan(2^n\cdot \alpha )$
• B. $cot\alpha -2^n\cdot cot(2^n\cdot \alpha )$
• C. $sec\alpha -2^n\cdot sec(2^n\cdot \alpha )$
• D. None of these

Answer 1

Answer: (B)

$cot\alpha -2^n\cdot cot(2^n\cdot \alpha )$

Let P(n): $tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +....+2^{n-1}tan\left(2^{n-1}\alpha \right)=cot\alpha -2^n\cdot cot(2^n\cdot \alpha )$ ..... (1)

Step I: For $n=1$,

L.H.S. of $\left(1\right)=tan\alpha$

$=cot\alpha -cot\alpha +tan\alpha =cot\alpha -(cot\alpha -tan\alpha )$

$=cot\alpha -(cot\alpha -\frac{1}{cot\alpha })$

$=cot\alpha -2\left(\frac{co{t}^2\alpha -1}{2cot\alpha }\right)$

$=cot\alpha -2\left(\frac{co{s}^2\alpha -{\sin }^2\alpha }{2cos\alpha \sin \alpha }\right)$

$=cot\alpha -2cot2\alpha$

$=R.H.S.$ of (1)

Therefore, $P\left(1\right)$ is true.

Step II: Assume it is true for $n=k$, then

$P\left(k\right):tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +....+2^{k-1}tan\left(2^{k-1}\alpha \right)$

$=cot\alpha -2^{k}cot\left(2^k\alpha \right)$

Step III: For $n=k+1$,

$P(k+1):tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +...+2^{k-1}tan\left(2^{k-1}\alpha \right)+2^{k}tan\left(2^k\alpha \right)=cot\alpha -2^{k+1}cot\left(2^{k+1}\alpha \right)$

$L.H.S.=tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +....+2^{k-1}tan\left(2^{k-1}\alpha \right)+2^{k}tan\left(2^k\alpha \right)$

$=cot\alpha -2^{k}cot\left(2^k\alpha \right)+2^{k}tan\left(2^k\alpha \right)$ (By assumption step)

$=cot\alpha -2^k\left(cot\right(2^k\alpha )-tan(2^k\alpha \left)\right)$

$=cot\alpha -2^k\cdot 2\left(\frac{co{t}^2\left(2^k\alpha \right)-1}{2cot\left(2^k\alpha \right)}\right)$

$=cot\alpha -2^{k+1}\cdot cot(2\cdot 2^k\alpha )$

$=cot\alpha -2^{k+1}\cdot cot\left(2^{k+1}\alpha \right)$

$=R.H.S.$

This show that the result is true for $n=k+1$.

Hence by the principle of mathematical induction, the result is true for all n $\in$ N.