MHB Is \( n \) Equal to 23 in the Given Trigonometric Product Equation?

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The equation \( (1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n \) leads to the determination of \( n \). By using the identity \( 1 + \tan(45^\circ - k) = \frac{2}{1 + \tan k} \), pairs of terms can be simplified to yield \( (1 + \tan k)(1 + \tan(45^\circ - k)) = 2 \). This results in 22 pairs contributing to the product, giving \( (1+\tan 1)(1+\tan 2)\cdots(1+\tan 44) = 2^{22} \). Including \( 1 + \tan 45 = 2 \) leads to the final product being \( 2^{23} \). Therefore, \( n \) is determined to be 23.
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Given that $$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n$$, find $n$.
 
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anemone said:
Given that $$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n$$, find $n$.

Note k in degrees and degree symbol not mentioned

tan (45-k)) = (tan 45- tank )/( 1+ tan 45 tan k)
= ( 1- tan k)/ (1 + tan k)
So 1+ tan (45-k) = ( 1+ tan k + 1 – tan k) / ( 1+ tan k) = 2/(1+ tan k)
Or (1 + tan (45-k))(1+ tan k) = 2

So (1 + tan 1) ( 1+ tan 44) = 2
(1+ tan 2)(1 + tan 43) = 2
( 1 + tan 22)(1+ tan 23) = 2

Hence (1+tan 1)( 1+ tan 2) … ( 1 + tan 44) = 2^22

As 1 + tan 45 = 2 so multiplying we get
(1+tan 1)( 1+ tan 2) … ( 1 + tan 44)( 1+ tan 45) = 2^23
hence n = 23
 
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