Solve tan - 1 2 x + tan - 1 3 x = π 4

We have tan - 1 2 x + tan - 1 3 x = π 4 or tan - 1 2 x + 3 x 1 - 2 x × 3 x = π 4 i.e. tan - 1 5 x 1 - 6 x 2 = π 4 Therefore 5 x 1 - 6 x 2 = tan π 4 = 1 or 6 x 2 + 5 x - 1 = 0 i.e., 6 x - 1 x + 1 = 0 Which gives x = 1 6 or x = - 1 . Since x = - 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative, x = 1 6 is the only solution of the given equation.

Solve tan - 1 2 x + tan - 1 3 x = π 4

We have tan - 1 2 x + tan - 1 3 x = π 4 or tan - 1 2 x + 3 x 1 - 2 x × 3 x = π 4 i.e. tan - 1 5 x 1 - 6 x 2 = π 4 Therefore 5 x 1 - 6 x 2 = tan π 4 = 1 or 6 x 2 + 5 x - 1 = 0 i.e., 6 x - 1 x + 1 = 0 Which gives x = 1 6 or x = - 1 . Since x = - 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative, x = 1 6 is the only solution of the given equation.

Prove that: 3 sin - 1 x = sin - 1 3 x - 4 x 3 , x ∈ - 1 2 , 1 2 .

To prove: 3 sin - 1 x = sin - 1 3 x - 4 x 3 , x ∈ - 1 2 , 1 2 Let sin - 1 x = θ ∴ x = sin θ Now, consider, RHS = sin - 1 ( 3 x - 4 x 3 ) = sin - 1 ( 3 sin θ - 4 sin 3 θ ) = sin - 1 ( sin 3 θ ) [Using: sin 3 θ = 3 sin θ - 4 sin 3 θ ] = 3 θ = 3 sin - 1 x = LHS Hence, proved.

Let f ( x ) = cos 2 tan - 1 sin cot - 1 1 - x x , 0 < x < 1 . Then:

( 1 - x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0

( 1 + x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0

( 1 - x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0

( 1 + x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0

If tan - 1 a + tan - 1 b + tan - 1 c = π , then prove that a + b + c = a b c

Given that: tan - 1 a + tan - 1 b + tan - 1 c = π To prove: a + b + c = a b c LHS can be written as tan - 1 a + tan - 1 b = π - tan - 1 c tan - 1 a + b 1 - a b = π - tan - 1 c ∵ tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y a + b 1 - a b = tan π - tan - 1 c Since, tan π - θ = - tan θ ∴ a + b 1 - a b = - tan tan - 1 c ⇒ a + b 1 - a b = - c ∵ tan tan - 1 x = x ⇒ a + b = - c + a b c ⇒ a + b + c = a b c Hence, proved.

Prove that: sin - 1 2 x 1 - x 2 = 2 cos - 1 x , 1 2 ≤ x ≤ 1 .

To prove: sin - 1 2 x 1 - x 2 = 2 cos - 1 x , 1 2 ≤ x ≤ 1 Put x = cos θ ⇔ θ = cos - 1 x Now, consider, LHS = sin - 1 2 x 1 - x 2 = sin - 1 2 cos θ 1 - cos 2 θ = sin - 1 ( 2 cos θ sin θ ) = sin - 1 ( sin 2 θ ) [Using: sin 2 θ = 2 sin θ cos θ ] = 2 θ = 2 cos - 1 x = RHS Hence, proved.