If sin - 1 ⁡ 2 a 1 + a 2 + sin - 1 ⁡ 2 b 1 + b 2 = 2 cot - 1 ⁡ 1 x , then x is equal to ∀ a , b ∈ 0 , 1

Question

If sin - 1 ⁡ 2 a 1 + a 2 + sin - 1 ⁡ 2 b 1 + b 2 = 2 cot - 1 ⁡ 1 x , then x is equal to ∀ a , b ∈ 0 , 1, a - b 1 + a b, b 1 + a b, b 1 - a b, a + b 1 - a b

MARKS App · Accepted Answer

Since, sin - 1 ⁡ 2 x 1 + x 2 = 2 tan - 1 ⁡ x for x ∈ - 1 , 1 s i n - 1 2 a 1 + a 2 + s i n - 1 2 b 1 + b 2 = 2 cot - 1 ⁡ 1 x ⇒ tan - 1 ⁡ a + tan - 1 ⁡ b = tan - 1 ⁡ x ⇒ tan - 1 ⁡ a + b 1 - a b = tan - 1 ⁡ x ⇒ x = a + b 1 - a b

Search any question & find its solution

Looking for more such questions to practice?