Let's begin with the left side:
tan(55∘)tan(65∘)tan(75∘)
Using the product to sum identity for the tangent function:
tan(α)tan(β)=cos(α−β)−cos(α+β)cos(α−β)−cos(α+β)
we may write:
tan(65∘)tan(55∘)=cos(10∘)−cos(120∘)cos(10∘)+cos(120∘)
Given that cos(120∘)=−12 we now have:
tan(65∘)tan(55∘)=cos(10∘)+12cos(10∘)−12=2cos(10∘)+12cos(10∘)−1
Hence, we may write:
tan(55∘)tan(65∘)tan(75∘)=2cos(10∘)+12cos(10∘)−1⋅tan(75∘)
Using the identity:
tan(α)=sin(α)cos(α)
we have:
tan(55∘)tan(65∘)tan(75∘)=(2cos(10∘)+1)sin(75∘)(2cos(10∘)−1)cos(75∘)
Distributing, there results:
tan(55∘)tan(65∘)tan(75∘)=2sin(75∘)cos(10∘)+sin(75∘)2cos(75∘)cos(10∘)−cos(75∘)
Using the product-to-sum identity:
2sin(α)cos(β)=sin(α+β)+sin(α−β)
we have:
2sin(75∘)cos(10∘)=sin(85∘)+sin(65∘)
Using the product-to-sum identity:
2cos(α)cos(β)=cos(α+β)+cos(α−β)
we have:
2cos(75∘)cos(10∘)=cos(85∘)+cos(65∘)
And now we may state:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)+sin(65∘)+sin(75∘)cos(85∘)+cos(65∘)−cos(75∘)
Rearranging, we have:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)+sin(75∘)+sin(65∘)cos(85∘)−(cos(75∘)−cos(65∘))
Using the sum-to-product identity:
sin(α)+sin(β)=2sin(α+β2)cos(α−β2)
we have:
sin(75∘)+sin(65∘)=2sin(70∘)cos(5∘)
Using the sum-to-product identity:
cos(α)−cos(β)=−2sin(α+β2)sin(α−β2)
we have:
−(cos(75∘)−cos(65∘))=2sin(70∘)sin(5∘)
Thus, we now have:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)+2sin(70∘)cos(5∘)cos(85∘)+2sin(70∘)sin(5∘)
Using the co-function identities:
cos(α)=sin(90∘−α)
sin(α)=cos(90∘−α)
we have:
cos(5∘)=sin(85∘)
sin(5∘)=cos(85∘)
and we now may write:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)+2sin(70∘)sin(85∘)cos(85∘)+2sin(70∘)cos(85∘)
Factoring, we get:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)(1+2sin(70∘))cos(85∘)(1+2sin(70∘))
Dividing out common factors, we now have:
tan(55∘)tan(65∘)tan(75∘)=sin(85∘)cos(85∘)
Using the identity:
tan(α)=sin(α)cos(α)
we have:
tan(55∘)tan(65∘)tan(75∘)=tan(85∘)
Shown as desired.
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