Trigonometric Functions JEE 2025: 3 Real Solutions Explained

Q: How many real solutions are there in the JEE Advanced 2025 inverse trigonometric equation?

There are exactly three real solutions. Using α = tan θ / 3 leads to a cubic equation after applying the correct identities based on the value of α. The cases for |α| > 1 lead to no real solutions, and the n = ±1 cases are eliminated by range considerations.

Q: How to avoid mistakes in inverse trigonometric functions for JEE Advanced 2025?

Always bound the RHS of the equation first to limit the possible values of n in the general solution. Use the substitution to linearise, apply the correct identity as per the range of α, and check all potential roots including those that appear cancelled after division.

Q: How should you prepare for inverse trigonometric equations before JEE Advanced 2025?

Introduce a linear substitution like α = tan θ / 3 to simplify the arguments. Always bound the range of the RHS before applying the general solution θ = arctan(3α) + nπ. Exhaust all cases for the identities and verify the valid roots by ensuring they fit within the principal ranges.

Trigonometric Functions JEE 2025: Inverse Equation Has Three Solutions

JEE Advanced 2025 Mathematics Trigonometric Functions Inverse trigonometric equations

By Founder, JEEnius - IIT Kanpur Alumni · Aug 23, 2026 · 3 min read

Hard 4 min target

The total number of real solutions of the equation

$θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$

Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\tan^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $(- \frac{π}{2}, \frac{π}{2})$ , respectively.

Show answerAnswer

C) $3$

Explanation

Let

$α = \frac{\tan θ}{3}$

Then

$\tan θ = 3 α$

The equation becomes

$θ = \tan^{- 1} (6 α) - \frac{1}{2} \sin^{- 1} (\frac{18 α}{9 + 9 α^{2}})$

$θ = \tan^{- 1} (6 α) - \frac{1}{2} \sin^{- 1} (\frac{2 α}{1 + α^{2}})$

Also, since $\tan θ = 3 α$ , we can write

$θ = \tan^{- 1} (3 α) + n π$

where $n$ is an integer.

Now the right-hand side lies in a bounded interval because $\tan^{- 1} (6 α) \in (- \frac{π}{2}, \frac{π}{2})$ and $\frac{1}{2} \sin^{- 1} (\frac{2 α}{1 + α^{2}}) \in [- \frac{π}{4}, \frac{π}{4}]$ . Hence only $n = - 1, 0, 1$ are possible. We check the feasible principal case first.

For $| α | \leq 1$ , use the standard identity

$\sin^{- 1} (\frac{2 α}{1 + α^{2}}) = 2 \tan^{- 1} α$

Therefore the equation for $n = 0$ becomes

$\tan^{- 1} (3 α) = \tan^{- 1} (6 α) - \tan^{- 1} α$

Taking tangent on both sides,

$3 α = \frac{6 α - α}{1 + 6 α^{2}}$

$3 α = \frac{5 α}{1 + 6 α^{2}}$

This gives

$3 α (1 + 6 α^{2}) = 5 α$

Hence

$α (18 α^{2} - 2) = 0$

$α = 0$

$18 α^{2} = 2$

Thus

$α^{2} = \frac{1}{9}$

Therefore

$α = \pm \frac{1}{3}$

All three values $0, \frac{1}{3}, - \frac{1}{3}$ lie in $[- 1, 1]$ , so they are valid.

Now check $α > 1$ . In this range,

$\sin^{- 1} (\frac{2 α}{1 + α^{2}}) = π - 2 \tan^{- 1} α$

The principal equation becomes

$\tan^{- 1} (3 α) = \tan^{- 1} (6 α) - \frac{π}{2} + \tan^{- 1} α$

This implies

$\tan^{- 1} (6 α) + \tan^{- 1} α - \tan^{- 1} (3 α) = \frac{π}{2}$

Taking tangent leads to the condition

$\frac{7 α}{1 - 6 α^{2}} = - \frac{1}{3 α}$

$21 α^{2} = 6 α^{2} - 1$

This gives

$15 α^{2} = - 1$

which is impossible for real $α$ . Hence no solution for $α > 1$ .

Similarly, for $α < - 1$ ,

$\sin^{- 1} (\frac{2 α}{1 + α^{2}}) = - π - 2 \tan^{- 1} α$

The same tangent condition again gives

$15 α^{2} = - 1$

which is impossible. So no solution for $α < - 1$ .

Finally, for $n = 1$ or $n = - 1$ , the left side $\tan^{- 1} (3 α) + n π$ lies outside the compatible sign/range of the right side in the corresponding intervals, so no additional solutions occur.

Therefore the only real solutions correspond to

$α = - \frac{1}{3}, 0, \frac{1}{3}$

Hence the total number of real solutions is

$3$

Correct option is C.

How many real solutions does the JEE Advanced 2025 inverse trigonometric equation have?

The equation has exactly three real solutions. It asks for the total number of real θ satisfying

θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ}) .

Principal ranges are [−π/2, π/2] for arcsin and (−π/2, π/2) for arctan. Four options were given: 1, 2, 3 or 5 real solutions. The correct choice is 3.

What is the official step-by-step solution for the JEE Advanced 2025 inverse trigonometric equation?

The official solution sets α = tan θ / 3 so tan θ = 3α and rewrites the equation as

θ = \tan^{- 1} (6 α) - \frac{1}{2} \sin^{- 1} (\frac{2 α}{1 + α^{2}}) .

It writes θ = arctan(3α) + nπ and argues the RHS bounded interval forces only n = −1, 0, 1 possible. The RHS lies between −3π/4 and 3π/4.

For |α| ≤ 1 use the identity arcsin(2α/(1+α²)) = 2 arctan α. The n = 0 case reduces to arctan(3α) = arctan(6α) − arctan α.

Taking tangent yields 3α = 5α / (1 + 6α²). Clearing the denominator produces 3α(1 + 6α²) = 5α then α(18α² − 2) = 0.

The roots α = 0, α = ±1/3 all lie inside [−1, 1] and are valid.

For α > 1 use arcsin(2α/(1+α²)) = π − 2 arctan α leading to arctan(6α) + arctan α − arctan(3α) = π/2. The tangent step produces 7α / (1 − 6α²) = −1/(3α) which simplifies to 15α² = −1, impossible for real α.

The identical contradiction appears for α < −1 with the −π − 2 arctan α form. The n = ±1 cases are ruled out by direct range mismatch with the RHS interval. Exactly three real solutions correspond to the three α values.

How does dividing by α produce the wrong answer of two solutions?

After reaching 3α(1 + 6α²) = 5α many students divide both sides by α assuming α ≠ 0. This cancels the α = 0 root and leaves only the quadratic 18α² − 2 = 0 giving α = ±1/3. The result is two solutions instead of three, selecting option B.

The error is failure to check the discarded root separately in the original equation before dividing. α = 0 satisfies the original equation because both sides become zero.

What range and identity rules must you keep ready for these equations?

Arcsin(2α/(1+α²)) equals 2 arctan α only when |α| ≤ 1. It equals π − 2 arctan α when α > 1 and −π − 2 arctan α when α < −1.

The RHS of the given equation is confined between −3π/4 and 3π/4 forcing n = −1, 0, 1 only. Tan θ = 3α implies the general solution θ = arctan(3α) + nπ for integer n. Comparison of intervals after substitution eliminates n = ±1.

Which two related questions recycle the same substitution and case analysis?

Find all real x satisfying

x = 2 \tan^{- 1} (3 x) - \frac{1}{2} \sin^{- 1} (\frac{6 x}{1 + 9 x^{2}})

and count the solutions. Solve

\tan^{- 1} (4 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{8 \tan θ}{4 + \tan^{2} θ}) = θ

for the number of real θ in (−π, π).

Both require identical α-substitution, identity split at |α| = 1 and nπ range check. Search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

How should you prepare for inverse trigonometric equations before JEE Advanced 2025?

Always introduce linear substitution of the form k tan θ to linearise the arguments inside inverse functions. Never apply the double-angle style identity for arcsin(2t/(1+t²)) outside |t| ≤ 1.

Explicitly bound the RHS before writing the general solution to limit possible n. Verify each candidate root by plugging back only after all algebraic cases are exhausted. This question’s 240-second expected time shows that careful case work beats speed.

If you get stuck on a similar question, photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

If that step was the hard part, work through Gravitation JEE 2024: h/H = 1/2 Maximises Total Time.

Frequently asked questions

How many real solutions are there in the JEE Advanced 2025 inverse trigonometric equation?

There are exactly three real solutions. Using α = tan θ / 3 leads to a cubic equation after applying the correct identities based on the value of α. The cases for |α| > 1 lead to no real solutions, and the n = ±1 cases are eliminated by range considerations.

Why do students get two solutions instead of three in trigonometric functions JEE 2025 problem?

Many students divide the equation 3α(1 + 6α²) = 5α by α and lose the α = 0 root. This root must be checked separately as it satisfies the original equation by making both sides zero. This leads to selecting the incorrect option of two solutions.

What is the correct identity for arcsin(2α/(1 + α^2)) when solving inverse trig equations?

The identity arcsin(2α/(1+α²)) = 2 arctan(α) holds only for |α| ≤ 1. For α > 1, it is π - 2 arctan(α), and for α < -1 it is -π - 2 arctan(α). These are essential to solve the equation correctly for all possible α.

How to avoid mistakes in inverse trigonometric functions for JEE Advanced 2025?

Always bound the RHS of the equation first to limit the possible values of n in the general solution. Use the substitution to linearise, apply the correct identity as per the range of α, and check all potential roots including those that appear cancelled after division.

How should you prepare for inverse trigonometric equations before JEE Advanced 2025?

Introduce a linear substitution like α = tan θ / 3 to simplify the arguments. Always bound the range of the RHS before applying the general solution θ = arctan(3α) + nπ. Exhaust all cases for the identities and verify the valid roots by ensuring they fit within the principal ranges.

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