To show: (1−tan2θ/2)(1−tan2θ/22)(1−tan2θ/23)...... =θ∗cotθ
or Equivalently; limn→∞(1−tan2θ/2)(1−tan2θ/22)... (1−tan2θ/2n)=θ∗cotθ
LHS =limn→∞(1−tan2θ/2)(1−tan2θ/22)...(1−tan2θ/2n)∗tanθ∗cotθ
=cotθ∗limn→∞(1−tan2θ/2)(1−tan2θ/22)... (1−tan2θ/2n)(2tanθ/2)/(1−tan2θ/2)
=cotθ∗limn→∞(1−tan2θ/22)...(1−tan2θ/2n) (2tanθ/2)
=cotθ∗limn→∞(1−tan2θ/22)...(1−tan2θ/2n) (22tanθ/22)/(1−tan2θ/22)
...
=cotθ∗limn→∞(1−tan2θ/2n) (2ntanθ/2n)/(1−tan2θ/2n)
=cotθ∗limn→∞2ntanθ/2n∗(θ/θ)
=θ∗cotθ∗limn→∞(tan(θ/2n))/(θ/2n)
=θcotθ
=RHS
Hence Proved.
Formulas used:
- tan2θ=2tanθ/(1−tan2θ)
- limx→0(tanx)/x=1
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