JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Evaluating the Sum of Inverse Trigonometric Expressions with Radicals

tan ^(-1)((1+√(3))/(3+√(3)))+sec ^(-1)(√((8+4 √(3))/(6+3 √(3)))) is equal to:

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Aπ/3Bπ/4Cπ/6Dπ/2

Step-by-step solutionView

Correct answer

Simplifying the arguments of both inverse trigonometric functions yields $\tan^{- 1} (1 / \sqrt{3}) + \sec^{- 1} (2 / \sqrt{3}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .

Option analysis

Why each option works or fails

A · π/3

The student correctly simplifies both radical fractions to evaluate the sum. Correct: $\tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ and $\sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6}$ , giving a sum of $\frac{π}{3}$ .

B · π/4

The student miscalculated the simplified value of the arguments, mistaking the first argument for $1$ (giving $\frac{π}{4}$ ) or adding the angles incorrectly. Factor $\sqrt{3}$ from the denominator $3 + \sqrt{3} = \sqrt{3} (\sqrt{3} + 1)$ to cancel the numerator, leaving $\frac{1}{\sqrt{3}}$ , not $1$ .

C · π/6

The student evaluated only the first term (or noted each individual angle is $\frac{π}{6}$ ) and stopped without adding them together. Remember to compute the full sum indicated by the plus sign: $\frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .

D · π/2

The student mistakenly assumed the identity $\tan^{- 1} (x) + \cot^{- 1} (x) = \frac{π}{2}$ applied directly by incorrectly equating $\sec^{- 1}$ to $\cot^{- 1}$ . Evaluate each inverse trigonometric value directly rather than forcing a complementary angle identity.

Reviewed route

Solution

StepWorking

01Given

We are given the expression $E = \tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ .

02Goal

Evaluate the exact value of $E$ in terms of radians.

03Approach

Factor out common surd factors from the numerators and denominators inside the inverse trigonometric functions to reduce them to standard values.

04Execute

For the first term: denominator is $3 + \sqrt{3} = \sqrt{3} (\sqrt{3} + 1)$ . Hence, $\frac{1 + \sqrt{3}}{3 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{3} (1 + \sqrt{3})} = \frac{1}{\sqrt{3}}$ . Thus, $\tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ .

05Execute

For the second term: inside the square root, factor $4$ from numerator and $3$ from denominator: $\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}} = \frac{4 (2 + \sqrt{3})}{3 (2 + \sqrt{3})} = \frac{4}{3}$ . Taking square root gives $\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ . Thus, $\sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6}$ .

06Execute

Sum the two angles: $E = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .

✓Verify

Since $\tan (π / 6) = 1 / \sqrt{3}$ and $\cos (π / 6) = \sqrt{3} / 2 ⟹ \sec (π / 6) = 2 / \sqrt{3}$ , both terms correctly evaluate to $π / 6$ . Sum is $π / 3$ .

Hints that build this answer step by step

What does the argument of the first term, $\frac{1 + \sqrt{3}}{3 + \sqrt{3}}$ , simplify to?
$\frac{1}{\sqrt{3}}$
What does the argument of the second term, $\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}}$ , simplify to?
$\frac{2}{\sqrt{3}}$
What is the value of $\tan^{- 1} (\frac{1}{\sqrt{3}}) + \sec^{- 1} (\frac{2}{\sqrt{3}})$ ?
$\frac{π}{3}$

✓ 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: 9 September 2026

Quick checks

Students also ask

Do I need to rationalize the denominator by multiplying by the conjugate?

No, directly factoring out √3 reveals the common binomial factor (1 + √3), which cancels out immediately.

Answer

Simplifying the arguments of both inverse trigonometric functions yields $\tan^{- 1} (1 / \sqrt{3}) + \sec^{- 1} (2 / \sqrt{3}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .

Why each option works or fails

A · correct: π/3 - The student correctly simplifies both radical fractions to evaluate the sum. Correct: $\tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ and $\sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6}$ , giving a sum of $\frac{π}{3}$ .
B: π/4 - The student miscalculated the simplified value of the arguments, mistaking the first argument for $1$ (giving $\frac{π}{4}$ ) or adding the angles incorrectly. Factor $\sqrt{3}$ from the denominator $3 + \sqrt{3} = \sqrt{3} (\sqrt{3} + 1)$ to cancel the numerator, leaving $\frac{1}{\sqrt{3}}$ , not $1$ .
C: π/6 - The student evaluated only the first term (or noted each individual angle is $\frac{π}{6}$ ) and stopped without adding them together. Remember to compute the full sum indicated by the plus sign: $\frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .
D: π/2 - The student mistakenly assumed the identity $\tan^{- 1} (x) + \cot^{- 1} (x) = \frac{π}{2}$ applied directly by incorrectly equating $\sec^{- 1}$ to $\cot^{- 1}$ . Evaluate each inverse trigonometric value directly rather than forcing a complementary angle identity.

Step-by-step solution

Given: We are given the expression $E = \tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ .
Goal: Evaluate the exact value of $E$ in terms of radians.
Approach: Factor out common surd factors from the numerators and denominators inside the inverse trigonometric functions to reduce them to standard values.
Execute: For the first term: denominator is $3 + \sqrt{3} = \sqrt{3} (\sqrt{3} + 1)$ . Hence, $\frac{1 + \sqrt{3}}{3 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{3} (1 + \sqrt{3})} = \frac{1}{\sqrt{3}}$ . Thus, $\tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ .
Execute: For the second term: inside the square root, factor $4$ from numerator and $3$ from denominator: $\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}} = \frac{4 (2 + \sqrt{3})}{3 (2 + \sqrt{3})} = \frac{4}{3}$ . Taking square root gives $\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ . Thus, $\sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6}$ .
Execute: Sum the two angles: $E = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .
Verify: Since $\tan (π / 6) = 1 / \sqrt{3}$ and $\cos (π / 6) = \sqrt{3} / 2 ⟹ \sec (π / 6) = 2 / \sqrt{3}$ , both terms correctly evaluate to $π / 6$ . Sum is $π / 3$ .

Evaluating the Sum of Inverse Trigonometric Expressions with Radicals

What feels right?

Simplifying the arguments of both inverse trigonometric functions yields $\tan^{- 1} (1 / \sqrt{3}) + \sec^{- 1} (2 / \sqrt{3}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

We think you should solve this next.

The same chapter, from another paper — Trigonometry, 2023

If the solution of the equation…

The same trap, in a different chapter — Integral Calculus, 2023

Students also ask

Evaluating the Sum of Inverse Trigonometric Expressions with Radicals

Simplifying the arguments of both inverse trigonometric functions yields tan⁡−1(1/3)+sec⁡−1(2/3)=π6+π6=π3tan−1(1/3​)+sec−1(2/3​)=6π​+6π​=3π​.

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

The same chapter, from another paper — Trigonometry, 2023

If the solution of the equation…

The same trap, in a different chapter — Integral Calculus, 2023

Students also ask

Simplifying the arguments of both inverse trigonometric functions yields $\tan^{- 1} (1 / \sqrt{3}) + \sec^{- 1} (2 / \sqrt{3}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$ .