Question:

Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>, \theta \in(0,2 \pi)$, holds If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

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When solving inequalities involving inverse trigonometric functions, ensure to use the principal values and properties of the functions.

Updated On: Mar 20, 2025

$\frac{\pi}{8}$
$\frac{\pi}{48}$
$\frac{\pi}{16}$
$\frac{\pi}{12}$

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The Correct Option is D

Solution and Explanation

sin^{- 1} sin θ - (\frac{π}{2} - sin^{- 1} sin θ) > 0

\Rightarrow sin^{- 1} sin θ > \frac{π}{4}

\Rightarrow sin θ > \frac{1}{2}

So, θ \in (\frac{π}{4}, \frac{3 π}{4})

θ \in (\frac{π}{4}, \frac{3 π}{4}) = (a, b)

\Rightarrow b - a = \frac{π}{2} = α - β

\Rightarrow β = α - \frac{π}{2}

\Rightarrow α x^{2} + β x + sin^{- 1} [(x - 3)^{2} + 1] + cos^{- 1} [(x - 3)^{2} + 1] = 0

x = 3, 9 α + 3 β + \frac{π}{2} + 0 = 0

\Rightarrow 9 α + 3 (α - \frac{π}{2}) + \frac{π}{2} = 0

\Rightarrow 12 α - π = 0

α = \frac{π}{12}

So, the correct option is (D) : $\frac{\pi}{12}$

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Concepts Used:

Properties of Inverse Trigonometric Functions

The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:

Property Set 1:

Sin⁻¹(x) = cosec⁻¹(1/x), x∈ [−1,1]−{0}
Cos⁻¹(x) = sec⁻¹(1/x), x ∈ [−1,1]−{0}
Tan⁻¹(x) = cot⁻¹(1/x), if x > 0 (or) cot⁻¹(1/x) −π, if x < 0
Cot⁻¹(x) = tan⁻¹(1/x), if x > 0 (or) tan⁻¹(1/x) + π, if x < 0

Property Set 2:

Sin⁻¹(−x) = −Sin⁻¹(x)
Tan⁻¹(−x) = −Tan⁻¹(x)
Cos⁻¹(−x) = π − Cos⁻¹(x)
Cosec⁻¹(−x) = − Cosec⁻¹(x)
Sec⁻¹(−x) = π − Sec⁻¹(x)
Cot⁻¹(−x) = π − Cot⁻¹(x)

Property Set 3:

Sin⁻¹(1/x) = cosec⁻¹x, x≥1 or x≤−1
Cos⁻¹(1/x) = sec⁻¹x, x≥1 or x≤−1
Tan⁻¹(1/x) = −π + cot⁻¹(x)

Property Set 4:

Sin⁻¹(cos θ) = π/2 − θ, if θ∈[0,π]
Cos⁻¹(sin θ) = π/2 − θ, if θ∈[−π/2, π/2]
Tan⁻¹(cot θ) = π/2 − θ, θ∈[0,π]
Cot⁻¹(tan θ) = π/2 − θ, θ∈[−π/2, π/2]
Sec⁻¹(cosec θ) = π/2 − θ, θ∈[−π/2, 0]∪[0, π/2]
Cosec⁻¹(sec θ) = π/2 − θ, θ∈[0,π]−{π/2}
Sin⁻¹(x) = cos⁻¹[√(1−x2)], 0≤x≤1 = −cos⁻¹[√(1−x2)], −1≤x<0

Property Set 5:

Sin⁻¹x + Cos⁻¹x = π/2
Tan⁻¹x + Cot⁻¹(x) = π/2
Sec⁻¹x + Cosec⁻¹x = π/2

Property Set 6:

If x, y > 0

Tan⁻¹x + Tan⁻¹y = π + tan⁻¹ (x+y/ 1-xy), if xy > 1

Tan⁻¹x + Tan⁻¹y = tan⁻¹ (x+y/ 1-xy), if xy < 1

If x, y < 0

Tan⁻¹x + Tan⁻¹y = tan⁻¹ (x+y/ 1-xy), if xy < 1

Tan⁻¹x + Tan⁻¹y = -π + tan⁻¹ (x+y/ 1-xy), if xy > 1

Property Set 7:

sin⁻¹(x) + sin⁻¹(y) = sin⁻¹[x√(1−y2)+ y√(1−x2)]
cos⁻¹x + cos⁻¹y = cos⁻¹[xy−√(1−x2)√(1−y2)]

Property Set 8:

sin⁻¹(sin x) = −π−π, if x∈[−3π/2, −π/2]

= x, if x∈[−π/2, π/2]

= π−x, if x∈[π/2, 3π/2]

=−2π+x, if x∈[3π/2, 5π/2] And so on.

cos⁻¹(cos x) = 2π+x, if x∈[−2π,−π]

= −x, ∈[−π,0]

= x, ∈[0,π]

= 2π−x, ∈[π,2π]

=−2π+x, ∈[2π,3π]

tan⁻¹(tan x) = π+x, x∈(−3π/2, −π/2)

= x, (−π/2, π/2)

= x−π, (π/2, 3π/2)

= x−2π, (3π/2, 5π/2)

Property Set 9: