Let α = 3sin^-1 ( 6 11 ) and β = 3cos^-1 ( 4 9 ), where inverse trigonometric functions take only the principal values. Given below are two…
JEE Main 2026 — Mathematics Trigonometry
2026mcqmedium
Let α=3sin−1(116) and β=3cos−1(94), 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:
Official previous-year question
Held on 8 Apr 2026 · Verified 6 Jul 2026.
Options
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
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Solution
We are given α=3sin−1(116) and β=3cos−1(94).
First, we find the range of α.
Since 21<116<21, we have sin−1(21)<sin−1(116)<sin−1(21).
⇒6π<sin−1(116)<4π
Multiplying by 3, we get 2π<α<43π.
Since α lies in the second quadrant, cos(α)<0. Thus, Statement II is true.
Next, we find the range of β.
Since 0<94<21 and cos−1(x) is a decreasing function, we have cos−1(21)<cos−1(94)<cos−1(0).
⇒3π<cos−1(94)<2π
Multiplying by 3, we get π<β<23π.
Thus, β lies in the third quadrant.
Now, we determine the sign of cos(α+β).
Adding the inequalities for α and β:
2π+π<α+β<43π+23π
⇒23π<α+β<49π
The angle (α+β) lies in the interval (23π,49π), which covers the fourth quadrant and a part of the first quadrant. In both of these regions, the cosine function is strictly positive.
Therefore, cos(α+β)>0. Thus, Statement I is true.
Both Statement I and Statement II are true.
Answer: Both Statement I and Statement II are true
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