Inverse Trigonometry — Identities and Addition Formulae

• Apply complementary angle identities to convert between $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, $\cot^{-1}$
• Use negative-argument identities to simplify expressions with negative inputs
• Apply the $\tan^{-1}x + \tan^{-1}y$ addition formula with correct case selection ($xy \lessgtr 1$)
• Use double-angle formulae to convert $2\tan^{-1}x$ to $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ form
• Convert $\sin^{-1}x$ to $\cos^{-1}$ or $\tan^{-1}$ form using right-triangle relationships
• Verify standard identities by substitution and cross-check against principal value constraints

Level 1 — Foundations

Complementary angle identities: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad x \in [-1,1]$ $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$ $\sec^{-1}x + \cosec^{-1}x = \frac{\pi}{2}, \quad |x| \geq 1$

Negative argument identities: $\sin^{-1}(-x) = -\sin^{-1}x \quad \text{(odd function)}$ $\cos^{-1}(-x) = \pi - \cos^{-1}x \quad \text{(NOT odd)}$ $\tan^{-1}(-x) = -\tan^{-1}x \quad \text{(odd function)}$ $\cot^{-1}(-x) = \pi - \cot^{-1}x$

Conversion identities (for $x \geq 0$): $\sin^{-1}x = \cos^{-1}\!\sqrt{1-x^2} = \tan^{-1}\!\frac{x}{\sqrt{1-x^2}}$ $\cos^{-1}x = \sin^{-1}\!\sqrt{1-x^2} = \tan^{-1}\!\frac{\sqrt{1-x^2}}{x}$

Level 2 — JEE Depth

Addition formula for $\tan^{-1}$:

$\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\!\dfrac{x+y}{1-xy} & \text{if } xy < 1 \\[6pt] \pi + \tan^{-1}\!\dfrac{x+y}{1-xy} & \text{if } xy > 1,\; x > 0 \\[6pt] -\pi + \tan^{-1}\!\dfrac{x+y}{1-xy} & \text{if } xy > 1,\; x < 0 \end{cases}$

Why the $\pm\pi$ correction? When $xy > 1$, both $x$ and $y$ are positive (or both negative), meaning $\tan^{-1}x + \tan^{-1}y > \pi/2$ (or $< -\pi/2$). The raw $\tan^{-1}$ formula gives a value in $(-\pi/2, \pi/2)$, so we add $\pi$ (positive case) or $-\pi$ (negative case) to land in the correct quadrant.

Subtraction formula: $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\frac{x-y}{1+xy} \quad \text{if } xy > -1$

Double angle formulae:

$2\tan^{-1}x = \sin^{-1}\!\frac{2x}{1+x^2}, \quad |x| \leq 1$ $2\tan^{-1}x = \cos^{-1}\!\frac{1-x^2}{1+x^2}, \quad x \geq 0$ $2\tan^{-1}x = \tan^{-1}\!\frac{2x}{1-x^2}, \quad |x| < 1$

Derivation sketch for $\sin^{-1}$ form: Let $\tan^{-1}x = \alpha$, so $\tan\alpha = x$. Then $\sin(2\alpha) = \frac{2\tan\alpha}{1+\tan^2\alpha} = \frac{2x}{1+x^2}$. So $2\alpha = \sin^{-1}\!\frac{2x}{1+x^2}$, i.e., $2\tan^{-1}x = \sin^{-1}\!\frac{2x}{1+x^2}$. Valid for $|x|\leq 1$ since $2\alpha = 2\tan^{-1}x \in [-\pi/2, \pi/2]$ only when $|x|\leq 1$.

Example 1: Prove that $\tan^{-1}\!\dfrac{1}{2} + \tan^{-1}\!\dfrac{1}{3} = \dfrac{\pi}{4}$.

Step 1 — Check the condition for the addition formula.
x = 1/2, y = 1/3.
xy = (1/2)(1/3) = 1/6 < 1. ✓ Use the first case.

Step 2 — Apply the formula.
tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 − 1/6))
                          = tan⁻¹((5/6)/(5/6))
                          = tan⁻¹(1)
                          = π/4  ✓

Step 3 — Verify result is in principal branch.
π/4 ∈ (−π/2, π/2) ✓. No correction needed.

Answer: tan⁻¹(1/2) + tan⁻¹(1/3) = π/4  □

Example 2: Evaluate $2\tan^{-1}\!\dfrac{1}{\sqrt{3}}$ using the double-angle formula and verify directly.

Method 1 — Double angle formula:
x = 1/√3. Check |x| ≤ 1: 1/√3 ≈ 0.577 ≤ 1. ✓

Using 2tan⁻¹x = tan⁻¹(2x/(1−x²)):
2x = 2/√3
1 − x² = 1 − 1/3 = 2/3
2x/(1−x²) = (2/√3)/(2/3) = (2/√3) × (3/2) = 3/√3 = √3

So 2tan⁻¹(1/√3) = tan⁻¹(√3) = π/3.

Method 2 — Direct calculation:
tan⁻¹(1/√3) = π/6   (standard value, since tan(π/6) = 1/√3)
2 × π/6 = π/3  ✓

Both methods agree: 2tan⁻¹(1/√3) = π/3.

• Q1: Simplify $\sin^{-1}(-\sqrt{3}/2) + \cos^{-1}(-\sqrt{3}/2)$.
• Q2: Evaluate $\tan^{-1}(2) + \tan^{-1}(3)$ (note: $xy = 6 > 1$, both positive).
• Q3: Express $\sin^{-1}(3/5)$ in the form $\tan^{-1}(\text{something})$ using the conversion identity.
• Q4: Use the double-angle formula to evaluate $2\tan^{-1}(1)$ in two ways: via $\tan^{-1}$ form and via $\sin^{-1}$ form; confirm they agree.
• Stretch: Q5: Prove the identity $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \pi/4$ by applying the addition formula twice and checking $xy < 1$ at each step.

NCERT Chapter 2 link: Section 2.3, Properties 3–7 (complementary, negative argument, addition formulae); worked examples 2.7–2.15.

Exam connections: JEE Main: identity-based evaluation (2–3 Qs/year); choosing the correct case in $\tan^{-1}$ addition formula. JEE Advanced: proofs of identities; multi-identity combination in equations.

Study strategy: Write each identity on a separate flashcard with one worked example. Drill the $\tan^{-1}$ addition formula daily for two weeks — it appears in nearly every JEE paper. For the double-angle formulae, always state which condition on $x$ applies before using the formula.

• Slider-based $\tan^{-1}$ addition explorer: input $x$ and $y$, watch the system automatically select the correct case ($xy < 1$, $xy > 1$ positive, $xy > 1$ negative) and display the result graphically.
• Complementary pair visualiser: show $\sin^{-1}x + \cos^{-1}x = \pi/2$ geometrically using a right triangle where both angles are labelled.
• AI mentor reflection: "Why does the $\tan^{-1}$ addition formula need a $\pi$ correction when $xy > 1$? Draw a diagram with two angles summing to more than $\pi/2$ to see why."

• "Use a right triangle with legs $x$ and $\sqrt{1-x^2}$ to derive $\sin^{-1}x + \cos^{-1}x = \pi/2$ geometrically — why does the triangle argument break down if $x$ is negative?"
• "If $\tan^{-1}x + \tan^{-1}y = \pi/4$, what constraint does this place on $x$ and $y$? Can you express $y$ as a function of $x$?"
• Stretch: "Generalise the $\tan^{-1}$ addition formula to three terms: derive an expression for $\tan^{-1}a + \tan^{-1}b + \tan^{-1}c$ by applying the two-term formula twice, and state all the conditions needed."

• 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 robot kinematics, the inverse tangent function (atan2) is used to find joint angles from end-effector positions; implement a simple 2D inverse-kinematics solver using the $\tan^{-1}$ addition formula to find the elbow angle of a two-link arm.
• Future Skill track: AI Mastery — Trigonometric identities underlie the Fourier transform used in signal processing and audio AI; show how the double-angle formula appears when expanding $\sin(2\omega t)$ in a Fourier series.
• Coding extension: Write a Python function arctan_add(x, y) that returns $\tan^{-1}x + \tan^{-1}y$ using the exact formula (including the $\pm\pi$ correction based on $xy$ and the sign of $x$), and verify it against math.atan(x) + math.atan(y) for 10 test cases.

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