Solution
We have (tan10 tan15 tan80 tan75)
= Where tan(90-θ)=cotθ hence tan80=tan(90-10)=cot 10
Where cotθ is the inverse of tanθ
= In the same way tan75=tan(90-15)=cot 15
= Where cot 10=(1/tan10) & cot 15=(1/tan15)
=(tan10 × tan15 × $$\frac{1}{tan10}$$×$$\frac{1}{tan15}$$)
= 1
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