Write the following function in the simplest form: tan - 1 3 a 2 x - x 3 a 3 - 3 a 2 x

tan - 1 3 a 2 x - x 3 a 3 - 3 a x 2 Divide by a 3 , = tan - 1 3 x a - x 3 a 3 1 - 3 x 2 a 2 = tan - 1 3 x a - x a 3 1 - 3 x a 2 Put x a = tan θ , θ = tan - 1 x a = tan - 1 3 tan θ - tan 3 θ 1 - 3 tan 2 θ = tan - 1 ( tan 3 θ ) = 3 θ = 3 tan - 1 x a

Write the following function in the simplest form: tan - 1 3 a 2 x - x 3 a 3 - 3 a 2 x

tan - 1 3 a 2 x - x 3 a 3 - 3 a x 2 Divide by a 3 , = tan - 1 3 x a - x 3 a 3 1 - 3 x 2 a 2 = tan - 1 3 x a - x a 3 1 - 3 x a 2 Put x a = tan θ , θ = tan - 1 x a = tan - 1 3 tan θ - tan 3 θ 1 - 3 tan 2 θ = tan - 1 ( tan 3 θ ) = 3 θ = 3 tan - 1 x a

Find the value of cos - 1 x + cos - 1 x 2 + 3 - 3 x 2 2 where 1 2 ≤ x ≤ 1 .

Given, cos - 1 x + cos - 1 x 2 + 3 - 3 x 2 2 Let x = cos θ , where 0 ≤ θ ≤ π 3 because 1 2 ≤ x ≤ 1 . Therefore, we have = cos - 1 cos θ + cos - 1 1 2 cos θ + 3 2 sin θ = cos - 1 cos θ + cos - 1 cos π 3 - θ ; ∵ 0 ≤ θ ≤ π 3 ⇒ π 3 ≥ π 3 - θ ≥ 0 Now, we know that cos - 1 cos y = y , y ∈ 0 , π = θ + π 3 - θ = π 3 .

Find the value of sin - 1 sin 3 π 5 .

We have to find the principal value of sin - 1 sin 3 π 5 Let y = sin - 1 sin 3 π 5 ⇒ y = sin - 1 sin π - 2 π 5 ⇒ y = sin - 1 sin 2 π 5 ∵ sin ( π - x ) = sin x We know that, sin - 1 ( sin x ) = x , x ∈ - π 2 , π 2 ∴ y = sin - 1 sin 2 π 5 = 2 π 5 .

Write the simplest form of tan - 1 cos x - sin x cos x + sin x , 0 < x < π 2 .

Given that, we have, tan - 1 cos x - sin x cos x + sin x = tan - 1 cos x 1 - sin x cos x cos x 1 + sin x cos x = tan - 1 1 - sin x cos x 1 + sin x cos x = tan - 1 1 - tan x 1 + tan x = tan - 1 tan π 4 - tan x 1 + tan π 4 · tan x = tan - 1 tan π 4 - x = π 4 - x [Using tan - 1 ( tan θ ) = θ ; θ ∈ - π 2 , π 2 ] Hence, the simplest form of tan - 1 cos x - sin x cos x + sin x is π 4 - x .

Write the principal value of cos - 1 cos - 16 π 15 .

Given, cos - 1 cos - 16 π 15 = cos - 1 cos 16 π 15 = cos - 1 cos 2 π - 14 π 15 = cos - 1 cos 14 π 15 We know that, cos - 1 ( cos x ) = x , when x ∈ [ 0 , π ] ∴ cos - 1 cos - 16 π 15 = 14 π 15

Show that 2 tan − 1 ⁡ x = sin − 1 ⁡ 2 x 1 + x 2 = cos − 1 ⁡ 1 − x 2 1 + x 2 .

Firstly we will prove 2 tan − 1 ⁡ x = sin − 1 ⁡ 2 x 1 + x 2 Let tan − 1 ⁡ x = θ ∴ x = tan ⁡ θ R.H.S = sin - 1 2 x 1 + x 2 = sin − 1 ⁡ 2 tan ⁡ θ 1 + tan 2 ⁡ θ = sin − 1 ⁡ 2 tan ⁡ θ sec 2 ⁡ θ ∵ 1 + tan 2 A = sec 2 A = sin − 1 ⁡ ( 2 sin ⁡ θ cos ⁡ θ ) = sin − 1 ⁡ ( sin ⁡ 2 θ ) ∵ sin 2 A = 2 sin A cos A = 2 θ ∵ sin - 1 sin A = A = 2 tan − 1 ⁡ x = L.H.S Now consider: R.H.S = cos − 1 ⁡ 1 − x 2 1 + x 2 = cos − 1 ⁡ 1 − tan 2 ⁡ θ 1 + tan 2 ⁡ θ = cos − 1 ⁡ cos 2 ⁡ θ − sin 2 ⁡ θ cos 2 ⁡ θ + sin 2 ⁡ θ = cos − 1 ⁡ ( cos ⁡ 2 θ ) ∵ cos 2 A = cos 2 A - sin 2 A , cos 2 A + sin 2 A = 1 = 2 θ = 2 tan - 1 x = L.H.S So, 2 tan − 1 ⁡ x = sin − 1 ⁡ 2 x 1 + x 2 = cos − 1 ⁡ 1 − x 2 1 + x 2 Hence, proved.

Prove that: tan - 1 1 + x 2 - 1 x = 1 2 tan - 1 x ; x ≠ 0

Let x = tan θ ⇒ θ = tan - 1 x Now, tan - 1 1 + x 2 - 1 x = tan - 1 1 + tan 2 θ - 1 tan θ = tan - 1 sec θ - 1 tan θ = tan - 1 1 cos θ - 1 sin θ cos θ = tan - 1 1 - cos θ sin θ = tan - 1 2 sin 2 θ 2 2 sin θ 2 cos θ 2 ∵ sin 2 θ = 2 sin θ cos θ ; cos 2 θ = 1 - 2 sin 2 θ = tan - 1 sin θ 2 cos θ 2 = tan - 1 tan θ 2 = θ 2 ∴ tan - 1 1 + x 2 - 1 x = 1 2 tan - 1 x .

Find the value of sin - 1 sin 3 π 5 .

Consider the given expression sin - 1 sin 3 π 5 L e t , x = sin - 1 sin 3 π 5 Now, the range of principal value of sin - 1 is - π 2 , π 2 ∴ x = 3 π 5 i s n o t p o s s i b l e So, sin x = sin 3 π 5 sin x = sin π - 3 π 5 ∴ sin x = sin 2 π 5 Since, the above value of x in the range of principal value Hence, sin - 1 sin 3 π 5 = 2 π 5

If A = sin - 1 2 x 1 + x 2 , B = cos - 1 1 - x 2 1 + x 2 and C = tan - 1 2 x 1 - x 2 satisfies the condition 3 A - 4 B + 2 C = π 3 . Find the value of x .

Given A = sin - 1 2 x 1 + x 2 , B = cos - 1 1 - x 2 1 + x 2 and C = tan - 1 2 x 1 - x 2 Given that they satisfy the condition 3 A - 4 B + 2 C = π 3 ⇒ 3 sin - 1 2 x 1 + x 2 - 4 cos - 1 1 - x 2 1 + x 2 + 2 tan - 1 2 x 1 - x 2 = π 3 Put x = tan y ⇒ 3 sin - 1 2 tan y 1 + tan 2 y - 4 cos - 1 1 - tan 2 y 1 + tan 2 y + 2 tan - 1 2 tan y 1 - tan 2 y = π 3 ⇒ 3 sin - 1 sin 2 y - 4 cos - 1 cos 2 y + 2 tan - 1 tan 2 y = π 3 ⇒ 3 2 y - 4 2 y + 2 2 y = π 3 ⇒ 6 y - 8 y + 4 y = π 3 ⇒ 2 y = π 3 ⇒ y = π 6 ⇒ tan - 1 x = π 6 ⇒ x = tan π 6 = 1 3 .

Find the principal value of sin - 1 sin 7 π 4 .

We know that sin - 1 sin x = x , x ∈ - π 2 , π 2 Given sin - 1 sin 7 π 4 = sin - 1 sin 2 π - π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 - sin π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 sin - π 4 ⇒ sin - 1 sin 7 π 4 = - π 4