If α=sin^(-1)⁡(√3)/2+sin^(-1)⁡1/3 and β=cos^(-1)⁡(√3)/2+cos^(-1)⁡1/3, then

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) α<β. α+β=sin^(-1)⁡(√3)/2+cos^(-1)⁡(√3)/2+sin^(-1)⁡1/3+cos^(-1)⁡1/3 =(π)/2+(π)/2=π Also, α=(π)/3+sin^(-1)⁡1/3<(π)/3+sin^(-1)⁡1/2 As sin⁡θ is increasing in 0, (π)/2 ∴ α<(π)/3+(π)/6=(π)/2 ⇒ β>(π)/2>α ⇒ α<β

If $α = \sin^{- 1} \frac{\sqrt{3}}{2} + \sin^{- 1} \frac{1}{3}$ and $β = \cos^{- 1} \frac{\sqrt{3}}{2} + \cos^{- 1} \frac{1}{3}$ , then

Solution

More Inverse Trigonometric Functions Questions

If α=sin-1⁡32+sin-1⁡13 and β=cos-1⁡32+cos-1⁡13, then

Solution

More Inverse Trigonometric Functions Questions

If $α = \sin^{- 1} \frac{\sqrt{3}}{2} + \sin^{- 1} \frac{1}{3}$ and $β = \cos^{- 1} \frac{\sqrt{3}}{2} + \cos^{- 1} \frac{1}{3}$ , then