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Inverse Trigonometric Functions question

2025 · Shift 2 · Q19
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  5. /2025 · Shift 2 · Q19

Inverse Trigonometric Functions question

2025 · Shift 2 · Q19

JEE AdvancedMathematicsInverse Trigonometric FunctionsMCQ+3 / −1
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)θ=tan−1(2tanθ)−21​sin−1(9+tan2θ6tanθ​) is (Here, the inverse trigonometric functions sin⁡−1x\sin^{-1} xsin−1x and tan⁡−1x\tan^{-1} xtan−1x assume values in [−π2,π2][ -\frac{\pi}{2}, \frac{\pi}{2}][−2π​,2π​] and (−π2,π2)( -\frac{\pi}{2}, \frac{\pi}{2})(−2π​,2π​), respectively.)
  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    5
View written solutionFree

Correct answer: C

  1. Let t=tan⁡θ.t=\tan\theta.t=tanθ. Since the RHS contains tan⁡θ\tan\thetatanθ, we must have cos⁡θ≠0\cos\theta\neq 0cosθ=0.

    The equation becomes θ=tan⁡−1(2t)−12sin⁡−1(6t9+t2).\theta=\tan^{-1}(2t)-\frac12\sin^{-1}\left(\frac{6t}{9+t^2}\right).θ=tan−1(2t)−21​sin−1(9+t26t​).

  2. Simplify the sin⁡−1\sin^{-1}sin−1 term

    Observe that 6t9+t2=2⋅3t9+t2.\frac{6t}{9+t^2}=\frac{2\cdot 3t}{9+t^2}.9+t26t​=9+t22⋅3t​.

    Now use the identity sin⁡2ϕ=2tan⁡ϕ1+tan⁡2ϕ.\sin 2\phi=\frac{2\tan\phi}{1+\tan^2\phi}.sin2ϕ=1+tan2ϕ2tanϕ​.

    Put tan⁡ϕ=t3.\tan\phi=\frac{t}{3}.tanϕ=3t​. Then sin⁡2ϕ=2(t/3)1+t2/9=6t9+t2.\sin 2\phi=\frac{2(t/3)}{1+t^2/9}=\frac{6t}{9+t^2}.sin2ϕ=1+t2/92(t/3)​=9+t26t​.

    Since ϕ=tan⁡−1(t/3)∈(−π2,π2)\phi=\tan^{-1}(t/3)\in\left(-\frac\pi2,\frac\pi2\right)ϕ=tan−1(t/3)∈(−2π​,2π​), we have 2ϕ∈(−π,π).2\phi\in(-\pi,\pi).2ϕ∈(−π,π). Therefore,

    \begin{cases} 2\phi, & |2\phi|\le \frac\pi2,\\[4pt] \pi-2\phi, & \frac\pi2<2\phi<\pi,\\[4pt] -\pi-2\phi, & -\pi<2\phi<- rac\pi2. \end{cases}$$ Now $|2\phi|\le \frac\pi2 \iff |\phi|\le \frac\pi4 \iff \left|\frac t3\right|\le 1 \iff |t|\le 3$. Hence $$\sin^{-1}\left(\frac{6t}{9+t^2}\right)= \begin{cases} 2\tan^{-1}(t/3), & |t|\le 3,\\[4pt] \pi-2\tan^{-1}(t/3), & t>3,\\[4pt] -\pi-2\tan^{-1}(t/3), & t<-3. \end{cases}$$

  3. Case-wise reduction

    The equation is θ=tan⁡−1(2t)−12sin⁡−1(6t9+t2).\theta=\tan^{-1}(2t)-\frac12\sin^{-1}\left(\frac{6t}{9+t^2}\right).θ=tan−1(2t)−21​sin−1(9+t26t​).

    Case 1: ∣t∣≤3|t|\le 3∣t∣≤3

    Then θ=tan⁡−1(2t)−tan⁡−1(t/3).\theta=\tan^{-1}(2t)-\tan^{-1}(t/3).θ=tan−1(2t)−tan−1(t/3).

    Use tan⁡−1a−tan⁡−1b=tan⁡−1(a−b1+ab)\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)tan−1a−tan−1b=tan−1(1+aba−b​) when the result lies in principal range. Here, 2t−t/31+2t2/3=5t3+2t2.\frac{2t-t/3}{1+2t^2/3}=\frac{5t}{3+2t^2}.1+2t2/32t−t/3​=3+2t25t​. So θ=tan⁡−1(5t3+2t2).\theta=\tan^{-1}\left(\frac{5t}{3+2t^2}\right).θ=tan−1(3+2t25t​).

    Taking tangent on both sides, t=5t3+2t2.t=\frac{5t}{3+2t^2}.t=3+2t25t​. Hence t(3+2t2)=5tt(3+2t^2)=5tt(3+2t2)=5t 2t3−2t=02t^3-2t=02t3−2t=0 2t(t2−1)=0.2t(t^2-1)=0.2t(t2−1)=0. Thus t=0, ±1.t=0,\ \pm 1.t=0, ±1.

    All satisfy ∣t∣≤3|t|\le 3∣t∣≤3.

    Corresponding θ\thetaθ values (since RHS is in principal range, these are unique): t=0⇒θ=0,t=0\Rightarrow \theta=0,t=0⇒θ=0, t=1⇒θ=π4,t=1\Rightarrow \theta=\frac\pi4,t=1⇒θ=4π​, t=−1⇒θ=−π4.t=-1\Rightarrow \theta=-\frac\pi4.t=−1⇒θ=−4π​.

    So this case gives 3 solutions.

    Case 2: t>3t>3t>3

    Then

    =\tan^{-1}(2t)+\tan^{-1}(t/3)-\frac\pi2.$$ Since $t>3$, both arctangents are positive. Let $$A=\tan^{-1}(2t),\quad B=\tan^{-1}(t/3).$$ Then $A,B\in(0,\pi/2)$ and $$\theta=A+B-\frac\pi2.$$ Taking tangent, $$t=\tan\left(A+B-\frac\pi2\right)=-\cot(A+B).$$ Now $$\tan(A+B)=\frac{2t+t/3}{1-2t^2/3}=\frac{7t/3}{(3-2t^2)/3}=\frac{7t}{3-2t^2}.$$ Hence $$-\cot(A+B)=-\frac{1}{\tan(A+B)}=-\frac{3-2t^2}{7t}=\frac{2t^2-3}{7t}.$$ So $$t=\frac{2t^2-3}{7t}.$$ Therefore $$7t^2=2t^2-3$$ $$5t^2=-3,$$ impossible for real $t$. So **no solution** for $t>3$. ### Case 3: $t<-3$ Then $$\theta=\tan^{-1}(2t)-\frac12\left(-\pi-2\tan^{-1}(t/3)\right) =\tan^{-1}(2t)+\tan^{-1}(t/3)+\frac\pi2.$$ Let again $$A=\tan^{-1}(2t),\quad B=\tan^{-1}(t/3).$$ Here $A,B<0$, so expression is valid. Taking tangent, $$t=\tan\left(A+B+\frac\pi2\right)=-\cot(A+B).$$ As before, $$\tan(A+B)=\frac{7t}{3-2t^2},$$ so $$t=-\frac{1}{\tan(A+B)}=\frac{2t^2-3}{7t}.$$ This again gives $$5t^2=-3,$$ impossible. Hence **no solution** for $t<-3$.

  4. Total number of real solutions

    Only Case 1 contributes, giving θ=−π4, 0, π4.\theta= -\frac\pi4,\ 0,\ \frac\pi4.θ=−4π​, 0, 4π​.

    Therefore, the total number of real solutions is 3.\boxed{3}.3​.

  5. Compare with stored answer

    Stored correct answer: C = 333.

    This matches our derived answer.

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