Solve the following equation for x : tan - 1 1 - x 1 + x - 1 2 tan - 1 x = 0 , where x > 0

Given that, tan - 1 1 - x 1 + x - 1 2 tan - 1 ( x ) = 0 ⇒ tan - 1 1 - x 1 + x = 1 2 tan - 1 ( x ) ⇒ tan - 1 1 - tan - 1 x = 1 2 tan - 1 ( x ) ∵ tan - 1 1 - tan - 1 x = tan - 1 1 - x 1 + x ⇒ tan - 1 1 = 3 2 tan - 1 ( x ) ⇒ π 4 = 3 2 tan - 1 ( x ) ⇒ π 6 = tan - 1 ( x ) ⇒ x = 1 3

Solve the following equation for x : cot - 1 x - cot - 1 ( x + 2 ) = π 12 , where x > 0

Given that, cot - 1 ( x ) - cot - 1 ( x + 2 ) = π 12 ⇒ tan - 1 1 x - tan - 1 1 x + 2 = π 12 ∵ cot - 1 x = tan - 1 1 x ⇒ tan - 1 1 x - 1 x + 2 1 + 1 x ( x + 2 ) = π 12 ⇒ tan - 1 2 x ( x + 2 ) x 2 + 2 x + 1 x ( x + 2 ) = π 12 ⇒ tan - 1 2 x 2 + 2 x + 1 = π 12 ⇒ 2 x 2 + 2 x + 1 = tan π 12 ⇒ 2 x 2 + 2 x + 1 = tan π 3 - π 4 ⇒ 2 x 2 + 2 x + 1 = tan π 3 - tan π 4 1 + tan π 3 × tan π 4 ⇒ 2 x 2 + 2 x + 1 = 3 - 1 3 + 1 ⇒ 2 x 2 + 2 x + 1 = 3 - 1 3 + 1 × 3 + 1 3 + 1 ⇒ 2 x 2 + 2 x + 1 = 2 ( 3 + 1 ) 2 ⇒ 1 ( x + 1 ) 2 = 1 ( 3 + 1 ) 2 ⇒ x + 1 = 3 + 1 ⇒ x = 3

Solve the following equation for x : tan - 1 x 2 + tan - 1 x 3 = π 4 , 0 < x < 6

Given that, tan - 1 x 2 + tan - 1 x 3 = π 4 , 0 < x < 6 We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y , x y < 1 ⇒ tan - 1 x 2 + x 3 1 - x 2 × x 3 = π 4 ⇒ tan - 1 5 x 6 6 - x 2 6 = π 4 ⇒ 5 x 6 - x 2 = tan π 4 ⇒ 5 x 6 - x 2 = 1 ⇒ x 2 + 5 x - 6 = 0 ⇒ ( x - 1 ) ( x + 6 ) = 0 ⇒ x = 1 [ ∵ 0 < x < 6 ]

Solve the following equation for x : tan - 1 x - 2 x - 4 + tan - 1 x + 2 x + 4 = π 4

We know tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y tan - 1 x - 2 x - 4 + tan - 1 x + 2 x + 4 = π 4 ⇒ tan - 1 x - 2 x - 4 + x + 2 x + 4 1 - x - 2 x - 4 × x + 2 x + 4 = π 4 ⇒ tan - 1 x 2 + 2 x - 8 + x 2 - 2 x - 8 ( x - 4 ) ( x + 4 ) x 2 - 16 - x 2 + 4 ( x - 4 ) ( x + 4 ) = π 4 ⇒ 2 x 2 - 16 - 12 = tan π 4 ⇒ 2 x 2 - 16 - 12 = 1 ⇒ 2 x 2 - 16 = - 12 ⇒ 2 x 2 = 4 ⇒ x 2 = 2 ⇒ x = ± 2

Solve the following equation for x : tan - 1 ( 2 + x ) + tan - 1 ( 2 - x ) = tan - 1 2 3 , where x < - 3 or, x > 3 .

Given that, tan - 1 ( 2 + x ) + tan - 1 ( 2 - x ) = tan - 1 2 3 We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y , x y < 1 ⇒ tan - 1 2 + x + 2 - x 1 - ( 2 + x ) × ( 2 - x ) = tan - 1 2 3 ⇒ 4 1 - 4 + x 2 = 2 3 ⇒ 2 x 2 = 18 ⇒ x 2 = 9 ⇒ x = ± 3

Solve the following equation for x : tan - 1 x - 2 x - 1 + tan - 1 x + 2 x + 1 = π 4

Given that, tan - 1 x - 2 x - 1 + tan - 1 x + 2 x + 1 = π 4 ⇒ tan - 1 x - 2 x - 1 + x + 2 x + 1 1 - x - 2 x - 1 x + 2 x + 1 = π 4 tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ⇒ ( x - 2 ) ( x + 1 ) + ( x - 1 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) ( x - 1 ) ( x + 1 ) - ( x - 2 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) = tan π 4 ⇒ ( x - 2 ) ( x + 1 ) + ( x - 1 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) - ( x - 2 ) ( x + 2 ) = 1 ⇒ x 2 - x - 2 + x 2 + x - 2 x 2 - 1 - x 2 - 4 = 1 ⇒ 2 x 2 - 4 3 = 1 ⇒ 2 x 2 - 4 = 3 ⇒ 2 x 2 = 7 ⇒ x 2 = 7 2 x = ± 7 2