Question #97342
Prove that
(1-tan^2(theta÷2))(1-tan^2(theta÷2square))(1-tan^2(theta÷2cube)).... Infinity= theta dot cot(theta)
1
Expert's answer
2019-10-31T06:26:12-0400

To show: (1tan2θ/2)(1tan2θ/22)(1tan2θ/23)......(1-tan^2\theta/2)(1-tan^2\theta/2^2)(1-tan^2\theta/2^3)...... =θcotθ=\theta*cot\theta


or Equivalently; limn(1tan2θ/2)(1tan2θ/22)...lim_{n \to \infty } (1-tan^2\theta/2)(1-tan^2\theta/2^2)... (1tan2θ/2n)=θcotθ(1-tan^2\theta/2^n)=\theta*cot\theta


LHS =limn(1tan2θ/2)(1tan2θ/22)...= lim_{n \to \infty}(1-tan^2\theta/2)(1-tan^2\theta/2^2)...(1tan2θ/2n)tanθcotθ(1-tan^2\theta/2^n)*tan\theta*cot\theta

=cotθlimn(1tan2θ/2)(1tan2θ/22)...=cot\theta *lim_{n \to \infty}\cancel{(1-tan^2\theta/2)}(1-tan^2\theta/2^2)... (1tan2θ/2n)(2tanθ/2)/(1tan2θ/2)(1-tan^2\theta/2^n)(2tan\theta/2)/\cancel{(1-tan^2\theta/2)}

=cotθlimn(1tan2θ/22)...(1tan2θ/2n)=cot\theta* lim_{n \to \infty}(1-tan^2\theta/2^2)...(1-tan^2\theta/2^n) (2tanθ/2)(2tan\theta/2)

=cotθlimn(1tan2θ/22)...(1tan2θ/2n)=cot\theta* lim_{n \to \infty}\cancel{(1-tan^2\theta/2^2)}...(1-tan^2\theta/2^n) (22tanθ/22)/(1tan2θ/22)(2^2tan\theta/2^2)/\cancel{(1-tan^2\theta/2^2)}

......

=cotθlimn(1tan2θ/2n)=cot\theta*lim_{n \to \infty}\cancel{(1-tan^2\theta/2^n)} (2ntanθ/2n)/(1tan2θ/2n)(2^ntan\theta/2^n)/\cancel{(1-tan^2\theta/2^n)}

=cotθlimn2ntanθ/2n(θ/θ)=cot\theta*lim_{n \to \infty}2^ntan\theta/2^n*(\theta/\theta)

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

=θcotθ=\theta cot\theta

=RHS=RHS


Hence Proved.


Formulas used:

  1. tan2θ=2tanθ/(1tan2θ)tan2\theta=2tan\theta/(1-tan^2\theta)
  2. limx0(tanx)/x=1lim_{x \to 0} (tanx)/x =1

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