Question

Applying the principal of mathematical induction of prove that
$\frac{2sin\theta }{cos\theta +3cos\theta }+\frac{2sin\theta }{cos\theta +5cos\theta }+\frac{2sin\theta }{cos\theta +cos(2n+1)\theta }$
= tan ( n + 1)$\theta -tan\theta$

Answer 1

$tanA-tanB=\frac{sin(A-B)}{cosAcosB}$
for P (1),
L.H.S. = $\frac{2sin\theta }{cos\theta +cos3\theta }=\frac{2sin\theta }{2cos\theta \cdot cos2\theta }$
$=\frac{sin(2\theta -\theta )}{cos\theta cos2\theta }=tan2\theta -tan\theta$
R.H.S. = $tan2\theta -tan\theta$ for n = 1
thus P(1) holds good. Assume P(n).
P(n + 1) = P(n) + $\frac{2sin\theta }{cos\theta +cos(2n+3)\theta }$
$=P\left(n\right)+\frac{2sin\left[\right(n+2)\theta -(n+1\left)\theta \right]}{2cos(n+2)\theta cos(n+1)\theta }$
$=\left[tan\right(n+1)\theta -tan\theta ]+\left[tan\right(n+2)\theta -tan(n+1\left)\theta \right]$
$=tan(n+2)\theta -tan\theta$
Hence P (n + 1) also holds good