Question

Cot60.tan45.tan30×tan75.tan15

Answer 1

$\bf\large\color{blue}\mathcal{HEY\:\:MATE}$
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Using,
Tan(90-A) = CotA
Cot(90-A) = TanA
TanA.CotA = 1
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Tan(90-60).Tan45.Tan30.Tan(90-75).Tan15

=> $Tan30.Tan45.\\Tan30.Cot15.Tan15$
=> $1 \times Tan^{2}30$
=> $Tan^{2}30°$
=>$\frac{1}{3}$
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Hope this helps!!!

Answer 2

To find: cot60° x tan45° x tan30° x tan75° x tan15°

=>cot60° x tan45° x tan30° x tan75° x tan15°=tan30° x tan45° x tan 30° x tan75° x cot75°                              [tan(90-θ)=cotθ and cot(90-θ)=tanθ]

=tan²30° x tan45° x 1         [cotθ * tanθ =1]

=(1/√3)² x 1

=1/3

Thanks!!!