JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Principal Values of Inverse Trigonometric Functions

Q: Why is cos^(-1)(cos(7π/6)) not equal to 7π/6 - π = π/6?

Because cos(7π/6) = -√(3)/2, which is negative. The principal branch of cos^(-1) is [0, π], where negative inputs yield angles in the second quadrant (π/2, π]. Hence, cos^(-1)(-√(3)/2) = π - π/6 = 5π/6.

sin ^(-1)(sin (2 π)/3)+cos ^(-1)(cos (7 π)/6)+tan ^(-1)(tan (3 π)/4) is equal to:

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A(11 π)/12B(17 π)/12C(31 π)/12D-(3 π)/4

Step-by-step solutionView

Correct answer

By evaluating each term within its principal value branch, we find $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{11 π}{12}$ .

Option analysis

Why each option works or fails

A · (11 π)/12

This is the correct response. Evaluating each term within the principal value ranges gives $\sin^{- 1} (\sin (2 π / 3)) = π / 3$ , $\cos^{- 1} (\cos (7 π / 6)) = 5 π / 6$ , and $\tan^{- 1} (\tan (3 π / 4)) = - π / 4$ . Adding them yields $\frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .

B · (17 π)/12

The student likely assumed $\tan^{- 1} (\tan (3 π / 4)) = \frac{π}{4}$ instead of $- \frac{π}{4}$ . Remember that $\tan (3 π / 4) = - 1$ , and the principal value branch of $\tan^{- 1} (x)$ is $(- π / 2, π / 2)$ , so $\tan^{- 1} (- 1) = - π / 4$ .

C · (31 π)/12

The student applied the identity $f^{- 1} (f (x)) = x$ directly to all three terms without checking the principal branch conditions. The relation $f^{- 1} (f (x)) = x$ only holds when $x$ lies in the principal value range of $f^{- 1}$ . For values outside this range, you must first find the equivalent angle within the range.

D · -(3 π)/4

The student committed sign and fraction addition errors after evaluating the individual inverse trigonometric terms. Ensure you correctly compute a common denominator for $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4}$ : using denominator 12 gives $\frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .

Reviewed route

Solution

StepWorking

01Given

Expression: $E = \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos \frac{7 π}{6}) + \tan^{- 1} (\tan \frac{3 π}{4})$ .

02Approach

Recall the principal value ranges: $\sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$ , $\cos^{- 1} x \in [0, π]$ , and $\tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})$ . Reduce each argument into its respective principal branch using periodicity and symmetry.

03Execute

For the first term: $\frac{2 π}{3} \notin [- \frac{π}{2}, \frac{π}{2}]$ . Since $\sin (\frac{2 π}{3}) = \sin (π - \frac{2 π}{3}) = \sin (\frac{π}{3})$ , we have $\sin^{- 1} (\sin \frac{2 π}{3}) = \frac{π}{3}$ . For the second term: $\frac{7 π}{6} \notin [0, π]$ . Since $\cos (\frac{7 π}{6}) = \cos (2 π - \frac{7 π}{6}) = \cos (\frac{5 π}{6})$ and $\frac{5 π}{6} \in [0, π]$ , we have $\cos^{- 1} (\cos \frac{7 π}{6}) = \frac{5 π}{6}$ . For the third term: $\frac{3 π}{4} \notin (- \frac{π}{2}, \frac{π}{2})$ . Since $\tan (\frac{3 π}{4}) = \tan (\frac{3 π}{4} - π) = \tan (- \frac{π}{4})$ , we have $\tan^{- 1} (\tan \frac{3 π}{4}) = - \frac{π}{4}$ .

04Execute

Sum the three evaluated values: $E = \frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .

✓Verify

Check individual numeric values: $\sin (2 π / 3) = \sqrt{3} / 2 ⟹ \sin^{- 1} (\sqrt{3} / 2) = π / 3$ . $\cos (7 π / 6) = - \sqrt{3} / 2 ⟹ \cos^{- 1} (- \sqrt{3} / 2) = 5 π / 6$ . $\tan (3 π / 4) = - 1 ⟹ \tan^{- 1} (- 1) = - π / 4$ . Sum: $4 π / 12 + 10 π / 12 - 3 π / 12 = 11 π / 12$ . Matches Option A.

Hints that build this answer step by step

What are the principal value branches for $\sin^{- 1} x$ , $\cos^{- 1} x$ , and $\tan^{- 1} x$ respectively?
$[- \frac{π}{2}, \frac{π}{2}]$ , $[0, π]$ , and $(- \frac{π}{2}, \frac{π}{2})$
What are the individual values of $\sin^{- 1} (\sin \frac{2 π}{3})$ , $\cos^{- 1} (\cos \frac{7 π}{6})$ , and $\tan^{- 1} (\tan \frac{3 π}{4})$ ?
$\frac{π}{3}$ , $\frac{5 π}{6}$ , and $- \frac{π}{4}$
What is the sum $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4}$ ?
$\frac{11 π}{12}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why is

\cos^{- 1} (\cos (7 π / 6))

not equal to

7 π / 6 - π = π / 6

Because $\cos (7 π / 6) = - \sqrt{3} / 2$ , which is negative. The principal branch of $\cos^{- 1}$ is $[0, π]$ , where negative inputs yield angles in the second quadrant $(π / 2, π]$ . Hence, $\cos^{- 1} (- \sqrt{3} / 2) = π - π / 6 = 5 π / 6$ .

Answer

By evaluating each term within its principal value branch, we find $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{11 π}{12}$ .

Why each option works or fails

A · correct: (11 π)/12 - This is the correct response. Evaluating each term within the principal value ranges gives $\sin^{- 1} (\sin (2 π / 3)) = π / 3$ , $\cos^{- 1} (\cos (7 π / 6)) = 5 π / 6$ , and $\tan^{- 1} (\tan (3 π / 4)) = - π / 4$ . Adding them yields $\frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .
B: (17 π)/12 - The student likely assumed $\tan^{- 1} (\tan (3 π / 4)) = \frac{π}{4}$ instead of $- \frac{π}{4}$ . Remember that $\tan (3 π / 4) = - 1$ , and the principal value branch of $\tan^{- 1} (x)$ is $(- π / 2, π / 2)$ , so $\tan^{- 1} (- 1) = - π / 4$ .
C: (31 π)/12 - The student applied the identity $f^{- 1} (f (x)) = x$ directly to all three terms without checking the principal branch conditions. The relation $f^{- 1} (f (x)) = x$ only holds when $x$ lies in the principal value range of $f^{- 1}$ . For values outside this range, you must first find the equivalent angle within the range.
D: -(3 π)/4 - The student committed sign and fraction addition errors after evaluating the individual inverse trigonometric terms. Ensure you correctly compute a common denominator for $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4}$ : using denominator 12 gives $\frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .

Step-by-step solution

Given: Expression: $E = \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos \frac{7 π}{6}) + \tan^{- 1} (\tan \frac{3 π}{4})$ .
Approach: Recall the principal value ranges: $\sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$ , $\cos^{- 1} x \in [0, π]$ , and $\tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})$ . Reduce each argument into its respective principal branch using periodicity and symmetry.
Execute: For the first term: $\frac{2 π}{3} \notin [- \frac{π}{2}, \frac{π}{2}]$ . Since $\sin (\frac{2 π}{3}) = \sin (π - \frac{2 π}{3}) = \sin (\frac{π}{3})$ , we have $\sin^{- 1} (\sin \frac{2 π}{3}) = \frac{π}{3}$ . For the second term: $\frac{7 π}{6} \notin [0, π]$ . Since $\cos (\frac{7 π}{6}) = \cos (2 π - \frac{7 π}{6}) = \cos (\frac{5 π}{6})$ and $\frac{5 π}{6} \in [0, π]$ , we have $\cos^{- 1} (\cos \frac{7 π}{6}) = \frac{5 π}{6}$ . For the third term: $\frac{3 π}{4} \notin (- \frac{π}{2}, \frac{π}{2})$ . Since $\tan (\frac{3 π}{4}) = \tan (\frac{3 π}{4} - π) = \tan (- \frac{π}{4})$ , we have $\tan^{- 1} (\tan \frac{3 π}{4}) = - \frac{π}{4}$ .
Execute: Sum the three evaluated values: $E = \frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}$ .
Verify: Check individual numeric values: $\sin (2 π / 3) = \sqrt{3} / 2 ⟹ \sin^{- 1} (\sqrt{3} / 2) = π / 3$ . $\cos (7 π / 6) = - \sqrt{3} / 2 ⟹ \cos^{- 1} (- \sqrt{3} / 2) = 5 π / 6$ . $\tan (3 π / 4) = - 1 ⟹ \tan^{- 1} (- 1) = - π / 4$ . Sum: $4 π / 12 + 10 π / 12 - 3 π / 12 = 11 π / 12$ . Matches Option A.

Principal Values of Inverse Trigonometric Functions

What feels right?

By evaluating each term within its principal value branch, we find $\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{11 π}{12}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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Principal Values of Inverse Trigonometric Functions

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