Let S 1 is the complete solution set of the inequality c o s - 1 x > c o s - 1 x 2 and S 2 is the complete solution set of the inequality c o t - 1 x 2 - 5 c o t - 1 x + 6 > 0 , then S 1 ∩ S 2 is

Step-by-step solution

c o s - 1 x > c o s - 1 x 2 ⇒ x < x 2 & x ∈ - 1 , 1 ∵ c o s - 1 x is a decreasing function ⇒ x x – 1 > 0 & x ∈ - 1,1 ⇒ x ∈ - ∞ , 0 ∪ 1 , ∞ & x ∈ - 1 , 1 ⇒ x ∈ - 1 , 0 … 1 Now c o t - 1 x 2 - 5 c o t - 1 x + 6 > 0 ⇒ c o t - 1 x - 2 c o t - 1 x - 3 > 0 ⇒ c o t - 1 x < 2 or x < c o t 3 … 2 ∵ c o t - 1 x is a decreasing function 1 ∩ 2 ⇒ c o t 2 , 0