JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Inverse Trigonometric Functions

Let

α = 3 sin^{- 1} (\frac{6}{11})

and

β = 3 cos^{- 1} (\frac{4}{9})

, where inverse trigonometric functions take only the principal values.

Given below are two statements:
Statement I :

cos (α + β) > 0

Statement II :

cos (α) < 0

In the light of the above statements, choose the correct answer from the options given below:

Answer: (A)

Solution

Since

\frac{1}{2} < \frac{6}{11} < \frac{1}{2}

sin^{- 1} \frac{6}{11} \in (\frac{π}{6}, \frac{π}{4})

, so

α = 3 sin^{- 1} \frac{6}{11} \in (\frac{π}{2}, \frac{3 π}{4})

, where

cos α < 0

; Statement II is true. Since

cos \frac{5 π}{12} < \frac{4}{9} < cos \frac{π}{3}

cos^{- 1} \frac{4}{9} \in (\frac{π}{3}, \frac{5 π}{12})

, so

β = 3 cos^{- 1} \frac{4}{9} \in (π, \frac{5 π}{4})

. Adding the two ranges,

α + β \in (\frac{3 π}{2}, 2 π)

, the fourth quadrant, where

cos (α + β) > 0

; Statement I is true. Both statements are true.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option