Inverse Trigonometry — Equations and JEE Problems

• Apply a systematic strategy to isolate and solve inverse trig equations
• Use complementary and negative-argument identities to reduce multi-term equations
• Recognise the three common JEE equation patterns and match the correct technique
• Apply substitution ($x = \sin\theta$ or $x = \tan\theta$) to simplify nested inverse trig expressions
• Form and solve quadratic equations that arise from inverse trig identities
• Validate solutions against domain constraints to reject extraneous roots

Level 1 — Foundations

General strategy for inverse trig equations:

1. Isolate one inverse trig term on one side if possible.
2. Apply the appropriate trig function to both sides.
3. Solve the resulting algebraic equation.
4. Check every solution in the original equation and confirm it lies in the valid domain.

Common JEE equation patterns:

Substitution technique: For expressions like $\sqrt{1-x^2}$ under $\sin^{-1}$ or $\cos^{-1}$: let $x = \sin\theta$. For expressions like $\frac{2x}{1+x^2}$: let $x = \tan\theta$.

This converts nested inverse trig to simple trig identities.

Addition formula for $\sin^{-1}$: $\sin^{-1}x + \sin^{-1}y = \sin^{-1}\!\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \quad \text{if } x^2+y^2 \leq 1$ $= \pi - \sin^{-1}\!\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \quad \text{if } x^2+y^2 > 1,\; x,y > 0$

Level 2 — JEE Depth

Solving $\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4$ — approach: Use the addition formula $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$ when $ab < 1$. Check: $ab = 6x^2$; need $6x^2 < 1$, i.e., $|x| < 1/\sqrt{6}$. Then: $\tan^{-1}\frac{5x}{1-6x^2} = \frac{\pi}{4}$ $\Rightarrow \frac{5x}{1-6x^2} = 1$ $\Rightarrow 6x^2 + 5x - 1 = 0$ $\Rightarrow (6x-1)(x+1) = 0$ $\Rightarrow x = 1/6$ or $x = -1$ (extraneous — check).

Domain-based rejection: $x = -1$ gives $\tan^{-1}(-2) + \tan^{-1}(-3) < 0 \neq \pi/4$. Reject. Only $x = 1/6$ is valid.

Solving $\sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x$ — approach: Use $\cos^{-1}x = \pi/2 - \sin^{-1}x$. $\sin^{-1}x + \sin^{-1}(1-x) = \pi/2 - \sin^{-1}x$ $2\sin^{-1}x + \sin^{-1}(1-x) = \pi/2$ $\sin^{-1}(1-x) = \pi/2 - 2\sin^{-1}x$ Take $\sin$ of both sides: $1 - x = \cos(2\sin^{-1}x) = 1 - 2x^2$ $2x^2 - x = 0$ $x(2x-1) = 0$ $x = 0$ or $x = 1/2$. Both must be checked in original equation (see Worked example 1).

Inequality in domain: For $\sin^{-1}x$ to exist: $x \in [-1,1]$. For $\sin^{-1}(1-x)$ to exist: $1-x \in [-1,1]$, i.e., $x \in [0,2]$. Combined: $x \in [0,1]$. Both $x=0$ and $x=1/2$ lie in $[0,1]$. ✓

Example 1: Solve $\sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x$.

Domain: x ∈ [0,1] (derived above).

Step 1 — Use complementary identity.
cos⁻¹x = π/2 − sin⁻¹x.
Equation becomes: sin⁻¹x + sin⁻¹(1−x) = π/2 − sin⁻¹x
⟹ 2sin⁻¹x = π/2 − sin⁻¹(1−x)
⟹ sin⁻¹(1−x) = π/2 − 2sin⁻¹x   … (*)

Step 2 — Apply sin to both sides.
sin(sin⁻¹(1−x)) = sin(π/2 − 2sin⁻¹x)
1 − x = cos(2sin⁻¹x)

Step 3 — Expand cos(2θ) where θ = sin⁻¹x.
cos(2θ) = 1 − 2sin²θ = 1 − 2x²
So: 1 − x = 1 − 2x²
⟹ 2x² − x = 0
⟹ x(2x − 1) = 0
⟹ x = 0 or x = 1/2.

Step 4 — Verify x = 0:
sin⁻¹(0) + sin⁻¹(1) = 0 + π/2 = π/2
cos⁻¹(0) = π/2 ✓

Step 5 — Verify x = 1/2:
sin⁻¹(1/2) + sin⁻¹(1/2) = π/6 + π/6 = π/3
cos⁻¹(1/2) = π/3 ✓

Both solutions are valid.
Answer: x = 0 and x = 1/2.

Example 2: Solve $\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}$.

Step 1 — Check condition for addition formula.
Let a = 2x, b = 3x. ab = 6x².
We need 6x² < 1 for the simple case: |x| < 1/√6.

Step 2 — Apply addition formula (assuming 6x² < 1 first):
tan⁻¹(2x) + tan⁻¹(3x) = tan⁻¹((2x + 3x)/(1 − 6x²)) = tan⁻¹(5x/(1 − 6x²))

Set equal to π/4:
tan⁻¹(5x/(1 − 6x²)) = π/4
⟹ 5x/(1 − 6x²) = tan(π/4) = 1
⟹ 5x = 1 − 6x²
⟹ 6x² + 5x − 1 = 0

Step 3 — Factorise:
Discriminant = 25 + 24 = 49
x = (−5 ± 7)/12
x = 2/12 = 1/6  or  x = −12/12 = −1

Step 4 — Check x = 1/6:
6x² = 6/36 = 1/6 < 1 ✓ (condition satisfied)
tan⁻¹(2/6) + tan⁻¹(3/6) = tan⁻¹(1/3) + tan⁻¹(1/2) = π/4 (verified by Example 1 of previous note) ✓

Step 5 — Check x = −1:
6x² = 6 > 1; condition violated. Our formula choice was invalid for this root.
Direct check: tan⁻¹(−2) + tan⁻¹(−3) = −tan⁻¹(2) − tan⁻¹(3)
= −(tan⁻¹(2) + tan⁻¹(3)) = −(π + tan⁻¹(5/(1−6))) [using xy>1, x>0 case after sign flip]
This gives a large negative value ≠ π/4. Reject x = −1.

Answer: x = 1/6.

• Q1: Solve $\sin^{-1}(2x) + \sin^{-1}(x) = \pi/3$ (form a quadratic in $x$ after applying the $\sin^{-1}$ addition formula or by taking $\sin$ of both sides strategically).
• Q2: Find all $x$ satisfying $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$. For what values of $x$ does this fail?
• Q3: Solve $\cos^{-1}x = 2\sin^{-1}x$. (Hint: use the complementary identity on $\cos^{-1}x$.)
• Q4: For what range of $x$ is $\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4$ valid to solve using the simple addition formula?
• Stretch: Q5: Solve $\sin^{-1}(1-x) - 2\sin^{-1}x = \pi/2$. Show all steps, identify the quadratic, and confirm the solution lies in the domain.

NCERT Chapter 2 link: Miscellaneous Exercise Q1–Q17 (equation-solving problems); worked examples 2.17–2.19.

Exam connections: JEE Main: 1–2 equation-solving problems per paper using complementary identity or $\tan^{-1}$ addition. JEE Advanced: multi-step problems requiring combination of identities, domain analysis, and quadratic solving; frequently a sub-part of a longer problem.

Study strategy: For every equation, write the domain constraint before solving — this alone eliminates extraneous solutions efficiently. Practice at least 15 varied equation-solving problems to recognise all JEE patterns. Time yourself: most one-step equations should take under 2 minutes.

• Equation solver visualiser: plot $y = \sin^{-1}x + \sin^{-1}(1-x)$ and $y = \cos^{-1}x$ on the same axes; the intersection points are the solutions — confirm they match the algebraic answer.
• Domain filter tool: input two inverse trig expressions and watch the system compute the intersection domain with a shaded number line.
• AI mentor reflection: "For the equation $\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4$, what changes if we replace $\pi/4$ with $3\pi/4$? Would there still be a valid solution?"

• "Why does squaring both sides or applying $\sin$ to both sides sometimes introduce extraneous roots in inverse trig equations, but the same operation in ordinary algebra does not?"
• "In Example 2, we rejected $x = -1$. Can you find a modified equation $\tan^{-1}(2x) + \tan^{-1}(3x) = c$ for which $x = -1$ would be a valid solution? What is $c$?"
• Stretch: "Solve the system: $\sin^{-1}x + \cos^{-1}y = \pi$ and $\tan^{-1}x - \tan^{-1}y = 0$. What does the geometric interpretation of each equation look like on the $xy$-plane?"

• Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
• 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
• Best real-world application stories (anonymised) featured on class or national leaderboard.

• In computer vision, inverse trig equations arise when solving for camera pan/tilt angles from detected object coordinates; simulate a simple "point the camera at object" problem using $\tan^{-1}$ and verify your solution programmatically.
• Future Skill track: AI Mastery — Iterative solvers (Newton's method) for transcendental equations use derivative information; compute the derivative of $\tan^{-1}(2x) + \tan^{-1}(3x)$ and implement one Newton's method iteration to find the root near $x = 0$.
• Coding extension: Write a Python script that uses scipy.optimize.brentq to numerically solve $\sin^{-1}x + \sin^{-1}(1-x) - \cos^{-1}x = 0$ over $[0,1]$, then verify both roots ($x=0$ and $x=0.5$) and plot the function to visualise the zeros.

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