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MathMedAdv 2025 • Paper 2

Total Number of Real Solutions to Inverse Trigonometric Equation

The total number of real solutions of the equation

θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left(\frac{6\tan\theta}{9+\tan^2\theta}\right)

is

(Here, the inverse trigonometric functions sin⁡−1x\sin^{-1} x and tan⁡−1x\tan^{-1} x assume values in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), respectively.)

Options

A

11

B

22

C

33

Correct
D

55

Topics & Concepts

Inverse Trigonometric FunctionsInverse Trigonometric FunctionsTrigonometric Equations

Step-by-Step Solution

To find the total number of real solutions to the equation θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left(\frac{6\tan\theta}{9+\tan^2\theta}\right)

Let t=tan⁡θt = \tan\theta. Note that the argument of the inverse sine function can be rewritten as: 6tan⁡θ9+tan⁡2θ=2(t3)1+(t3)2\frac{6\tan\theta}{9+\tan^2\theta} = \frac{2\left(\frac{t}{3}\right)}{1 + \left(\frac{t}{3}\right)^2}

Using the identity for sin⁡−1(2x1+x2)\sin^{-1}\left(\frac{2x}{1+x^2}\right):

  1. Case 1: ∣t∣≤3|t| \le 3 (i.e., ∣t3∣≤1\left|\frac{t}{3}\right| \le 1)

In this interval, sin⁡−1(2(t/3)1+(t/3)2)=2tan⁡−1(t3)\sin^{-1}\left(\frac{2(t/3)}{1+(t/3)^2}\right) = 2\tan^{-1}\left(\frac{t}{3}\right).

Substituting this into the given equation yields: θ=tan⁡−1(2t)−tan⁡−1(t3)\theta = \tan^{-1}(2t) - \tan^{-1}\left(\frac{t}{3}\right)

Taking the tangent of both sides: tan⁡θ=tan⁡(tan⁡−1(2t)−tan⁡−1(t3))\tan\theta = \tan\left(\tan^{-1}(2t) - \tan^{-1}\left(\frac{t}{3}\right)\right) t=2t−t31+2t(t3)=5t3+2t2t = \frac{2t - \frac{t}{3}}{1 + 2t\left(\frac{t}{3}\right)} = \frac{5t}{3 + 2t^2}

Rearranging the terms: t(3+2t2)−5t=0  ⟹  t(2t2−2)=0  ⟹  2t(t−1)(t+1)=0t(3 + 2t^2) - 5t = 0 \implies t(2t^2 - 2) = 0 \implies 2t(t - 1)(t + 1) = 0

This gives the roots t=0,1,−1t = 0, 1, -1, all of which lie within the interval [−3,3][-3, 3]:

  • For t=0t = 0: θ=0\theta = 0, which gives LHS=0=RHS\text{LHS} = 0 = \text{RHS}.
  • For t=1t = 1: θ=π4\theta = \frac{\pi}{4}, which gives LHS=π4\text{LHS} = \frac{\pi}{4} and RHS=tan⁡−1(2)−tan⁡−1(13)=tan⁡−1(1)=π4\text{RHS} = \tan^{-1}(2) - \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(1) = \frac{\pi}{4}.
  • For t=−1t = -1: θ=−π4\theta = -\frac{\pi}{4}, which gives LHS=−π4=RHS\text{LHS} = -\frac{\pi}{4} = \text{RHS}.

Any other values of θ\theta satisfying tan⁡θ=0,±1\tan\theta = 0, \pm 1 (where θ=kπ\theta = k\pi or θ=kπ±π4\theta = k\pi \pm \frac{\pi}{4} for k∈Z∖{0}k \in \mathbb{Z} \setminus \{0\}) do not satisfy the equation because the right-hand side is bounded strictly within (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

  1. Case 2: ∣t∣>3|t| > 3
  • For t>3t > 3: 12sin⁡−1(6t9+t2)=π2−tan⁡−1(t3)\frac{1}{2}\sin^{-1}\left(\frac{6t}{9+t^2}\right) = \frac{\pi}{2} - \tan^{-1}\left(\frac{t}{3}\right), leading to RHS=tan⁡−1(2t)+tan⁡−1(t3)−π2<0\text{RHS} = \tan^{-1}(2t) + \tan^{-1}\left(\frac{t}{3}\right) - \frac{\pi}{2} < 0, while θ>tan⁡−1(3)>0\theta > \tan^{-1}(3) > 0 for principal values, yielding no solutions.
  • For t<−3t < -3: by symmetry, no solutions exist.

Therefore, the only real solutions are θ=−π4,0,π4\theta = -\frac{\pi}{4}, 0, \frac{\pi}{4}.

The total number of real solutions is 33.

Correct Option: (C)

Total Number of Real Solutions to Inverse Trigonometric Equation | Mathematics PYQ Solution - JEE Challenger