Answer to Question #97342 in Trigonometry for Ojugbele Daniel

Question

Prove that
(1-tan^2(theta÷2))(1-tan^2(theta÷2square))(1-tan^2(theta÷2cube)).... Infinity= theta dot cot(theta)

Answer 1

To show: $(1-tan^2\theta/2)(1-tan^2\theta/2^2)(1-tan^2\theta/2^3)......$ $=\theta*cot\theta$

or Equivalently; $lim_{n \to \infty } (1-tan^2\theta/2)(1-tan^2\theta/2^2)...$ $(1-tan^2\theta/2^n)=\theta*cot\theta$

LHS $= lim_{n \to \infty}(1-tan^2\theta/2)(1-tan^2\theta/2^2)...$$(1-tan^2\theta/2^n)*tan\theta*cot\theta$

$=cot\theta *lim_{n \to \infty}\cancel{(1-tan^2\theta/2)}(1-tan^2\theta/2^2)...$ $(1-tan^2\theta/2^n)(2tan\theta/2)/\cancel{(1-tan^2\theta/2)}$

$=cot\theta* lim_{n \to \infty}(1-tan^2\theta/2^2)...(1-tan^2\theta/2^n)$ $(2tan\theta/2)$

$=cot\theta* lim_{n \to \infty}\cancel{(1-tan^2\theta/2^2)}...(1-tan^2\theta/2^n)$ $(2^2tan\theta/2^2)/\cancel{(1-tan^2\theta/2^2)}$

$...$

$=cot\theta*lim_{n \to \infty}\cancel{(1-tan^2\theta/2^n)}$ $(2^ntan\theta/2^n)/\cancel{(1-tan^2\theta/2^n)}$

$=cot\theta*lim_{n \to \infty}2^ntan\theta/2^n*(\theta/\theta)$

$=\theta*cot\theta*lim_{n \to \infty}(tan(\theta/2^n))/(\theta/2^n)$

$=\theta cot\theta$

$=RHS$

Hence Proved.

Formulas used:

1. $tan2\theta=2tan\theta/(1-tan^2\theta)$
2. $lim_{x \to 0} (tanx)/x =1$