JEE Main 2026 Apr 8 Shift 2, Mathematics Q12: Inverse Trigonometric Functions

JEE Main2026Apr 8, Shift 2Mathematics

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:

A
Both Statement I and Statement II are true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true

Solution

Bounding $α$ : Since $\frac{1}{2} < \frac{6}{11} < \frac{1}{2}$ ,

$sin^{- 1} \frac{1}{2} < sin^{- 1} \frac{6}{11} < sin^{- 1} \frac{1}{2} ⟹ \frac{π}{6} < sin^{- 1} \frac{6}{11} < \frac{π}{4}$

Multiplying by $3$ : $\frac{π}{2} < α < \frac{3 π}{4}$ . Hence $cos α < 0$ . Statement II is true.

Bounding $β$ : Since $0 < \frac{4}{9} < \frac{1}{2}$ and $cos^{- 1}$ is decreasing,

$\frac{π}{3} < cos^{- 1} \frac{4}{9} < \frac{π}{2}$

Multiplying by $3$ : $π < β < \frac{3 π}{2}$ .

Adding the bounds: $\frac{3 π}{2} < α + β < \frac{9 π}{4}$ .

This interval lies entirely in the fourth quadrant and a sliver past $2 π$ (i.e. equivalent to $(- \frac{π}{2}, \frac{π}{4})$ mod $2 π$ ), where cosine is positive. Hence $cos (α + β) > 0$ . Statement I is true.

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q12: Inverse Trigonometric Functions

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