Given, tan 2A = cot (A – 18°)
⇒ tan 2A = tan (90 – (A – 18°)
⇒ tan 2A = tan (108° – A)
⇒ 2A = 108° – A
⇒ 3A = 108°
⇒ A = 36°
Given, tan 4θ=cot(θ−10°)
This can be written as
cot(90°− 4θ)=cot(θ−10°) —–(i)
(∵ Tan θ = Cot(90°− θ))
Hence, from (i) we have
⇒90°− 4θ= θ−10°
⇒5θ =100°
⇒θ =20°
1/ sec θ = cos θ. And value of cos θ ranges from 0 to 1
Since cos 90° = 0
The given expression
cos 1° × cos 2° × cos 3° ×….× cos 90° ×……..× cos 180°
reduces to zero as it contains cos 90° which is equal to 0