JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Trigonometry: JEE Main Mathematics Question with Solution

\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)

is equal to :

Your answer stays private

What feels right?

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

\frac{\pi}{3}

\frac{\pi}{4}

\frac{\pi}{6}

\frac{\pi}{2}

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\frac{\pi}{3}

None. The student correctly simplifies both radical fractions to evaluate the sum. Correct: $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$ , giving a sum of $\frac{\pi}{3}$ .

B ·

\frac{\pi}{4}

The student miscalculated the simplified value of the arguments, mistaking the first argument for $1$ (giving $\frac{\pi}{4}$ ) 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 ·

\frac{\pi}{6}

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

D ·

\frac{\pi}{2}

The student mistakenly assumed the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$ 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 $E = \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$ .

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, $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ .

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, $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$ .

06execute

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

✓verify

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

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 $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) + \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ ?
$\frac{\pi}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\tan^{-1}(1/\sqrt{3}) + \sec^{-1}(2/\sqrt{3}) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ .

Why each option works or fails

A · correct: $\frac{\pi}{3}$ - None. The student correctly simplifies both radical fractions to evaluate the sum. Correct: $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$ , giving a sum of $\frac{\pi}{3}$ .
B: $\frac{\pi}{4}$ - The student miscalculated the simplified value of the arguments, mistaking the first argument for $1$ (giving $\frac{\pi}{4}$ ) or adding the angles incorrectly. Factor $\frac{\pi}{4}$ from the denominator $3+\sqrt{3} = \sqrt{3}(\sqrt{3}+1)$ to cancel the numerator, leaving $\frac{1}{\sqrt{3}}$ , not $1$ .
C: $\frac{\pi}{6}$ - The student evaluated only the first term (or noted each individual angle is $\frac{\pi}{6}$ ) and stopped without adding them together. Remember to compute the full sum indicated by the plus sign: $\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ .
D: $\frac{\pi}{2}$ - The student mistakenly assumed the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$ 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 $E = \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$ .
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, $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ .
execute: For the second term: inside the square root, factor $\cot^{-1}$ 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, $3+\sqrt{3} = \sqrt{3}(\sqrt{3}+1)$ .
execute: Sum the two angles: $E = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ .
verify: Since $\tan(\pi/6) = 1/\sqrt{3}$ and $\cos(\pi/6) = \sqrt{3}/2 \implies \sec(\pi/6) = 2/\sqrt{3}$ , both terms correctly evaluate to $\pi/6$ . Sum is $\pi/3$ .

Trigonometry: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

We think you should solve this next

Eight persons are to be transported from city A to city B in three cars of different make…

If the equation of the plane containing the line…

Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and…

Students also ask

Trigonometry: JEE Main Mathematics Question with Solution

Simplifying the arguments of both inverse trigonometric functions yields tan⁡−1(1/3)+sec⁡−1(2/3)=π6+π6=π3\tan^{-1}(1/\sqrt{3}) + \sec^{-1}(2/\sqrt{3}) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}tan−1(1/3​)+sec−1(2/3​)=6π​+6π​=3π​.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Eight persons are to be transported from city A to city B in three cars of different make…

If the equation of the plane containing the line…

Let f(x)+2f(1x)=x2+5f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5f(x)+2f(x1​)=x2+5 and…

Students also ask

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

Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and…