The Identity Toolkit

Why this section exists. The rationalized NCERT trimmed the classical identity apparatus of this chapter — the negative-argument rules, the reciprocal conversions, the complementary pairs and the addition formulas. JEE Main kept every one of them, and most JEE questions from this chapter are unsolvable without them. Board-only students may treat this as enrichment; JEE aspirants should treat it as core.

Negative arguments (read off the branch geometry)

The odd trio reflects through the origin; the $[0, \pi]$ family reflects through $\frac{\pi}{2}$:

$$\sin^{-1}(-x) = -\sin^{-1}x \qquad \tan^{-1}(-x) = -\tan^{-1}x \qquad \mathrm{cosec}^{-1}(-x) = -\mathrm{cosec}^{-1}x$$ $$\cos^{-1}(-x) = \pi - \cos^{-1}x \qquad \cot^{-1}(-x) = \pi - \cot^{-1}x \qquad \sec^{-1}(-x) = \pi - \sec^{-1}x$$

These formalise the two negative-input rules used throughout Sections 1 and 2.

Reciprocal conversions

$$\sin^{-1}\frac{1}{x} = \mathrm{cosec}^{-1}x, \qquad \cos^{-1}\frac{1}{x} = \sec^{-1}x \qquad (\vert x \vert \geq 1)$$ $$\tan^{-1}\frac{1}{x} = \cot^{-1}x \ (x > 0); \qquad \tan^{-1}\frac{1}{x} = \cot^{-1}x - \pi \ (x < 0)$$

The $x < 0$ correction in the last line is a JEE favourite: for negative $x$, $\ \tan^{-1}\frac{1}{x}$ is negative while $\cot^{-1}x$ sits in the second quadrant — they cannot be equal, and the $-\pi$ reconciles them.

The complementary pairs

$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \quad (x \in [-1, 1])$$ $$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \quad (x \in \mathbb{R}) \qquad \sec^{-1}x + \mathrm{cosec}^{-1}x = \frac{\pi}{2} \quad (\vert x \vert \geq 1)$$

Why (first pair): if $\theta = \sin^{-1}x$ then $\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta = x$, and $\frac{\pi}{2} - \theta \in [0, \pi]$, so $\cos^{-1}x = \frac{\pi}{2} - \theta$. These pairs power every "find $\cos^{-1}x$ given $\sin^{-1}x$" one-liner and many equation tricks (e.g. $\sin^{-1}x = \cos^{-1}x$ forces both to be $\frac{\pi}{4}$).

Addition Formulas and the $2\tan^{-1}x$ Trio

The tangent addition formulas

$$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy} \quad (xy < 1)$$ $$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x - y}{1 + xy} \quad (xy > -1)$$

The correction term (the JEE trap): if $x, y > 0$ with $xy > 1$, the true sum exceeds $\frac{\pi}{2}$ while the right side's principal value goes negative — add $\pi$: $$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy} \quad (x, y > 0, \ xy > 1)$$

Showcase: $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi$. Indeed $\tan^{-1}2 + \tan^{-1}3$ has $xy = 6 > 1$, so it equals $\pi + \tan^{-1}\frac{5}{-5} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$; adding $\tan^{-1}1 = \frac{\pi}{4}$ gives $\pi$.

The $2\tan^{-1}x$ trio (memorise all three)

$$2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2} \quad (\vert x \vert \leq 1)$$ $$2\tan^{-1}x = \cos^{-1}\frac{1 - x^2}{1 + x^2} \quad (x \geq 0)$$ $$2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2} \quad (\vert x \vert < 1)$$

All three are the $t$-substitution formulas ($\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$ in terms of $t = \tan\theta$) read backwards. Spotting the shapes $\frac{2x}{1+x^2}$, $\frac{1-x^2}{1+x^2}$, $\frac{2x}{1-x^2}$ inside an inverse function and collapsing them to $2\tan^{-1}x$ is the single most useful reflex in JEE questions on this chapter.

The symmetric-sum condition

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{\pi}{2}$, then $xy + yz + zx = 1$ — and if the sum is $\pi$, then $x + y + z = xyz$. (Both drop out of expanding $\tan$ of the sum; they appear in JEE as "which relation must hold" questions.)

Common mistakes to avoid

Mistake 1 — using the addition formula blind when $xy > 1$: the $\pi$-correction is mandatory; without it $\tan^{-1}2 + \tan^{-1}3 = -\frac{\pi}{4}$, absurd for a sum of two positive angles.

Mistake 2 — the trio's windows: $\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x$ fails for $\vert x \vert > 1$ (the left side is trapped in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; the correct value there is $\pi - 2\tan^{-1}x$ for $x > 1$).

Mistake 3 — solving equations without rejecting roots: clearing $\tan^{-1}$ from an equation squares away sign information; substitute every root back (see the worked equation below).

Mistake 4 — $\tan^{-1}\frac{1}{x} = \cot^{-1}x$ for negative $x$: false without the $-\pi$ correction.

JEE-Pattern Worked Examples

Example 1 — Complementary one-liner

If $\sin^{-1}x = \dfrac{\pi}{5}$ for some $x \in (-1, 1)$, find $\cos^{-1}x$.

Step 1 — the pair identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.

Step 2 — solve: $\cos^{-1}x = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}$.

Answer: $\dfrac{3\pi}{10}$ — no need to find $x$ at all.

Example 2 — The classic quarter-circle sum

Show that $\tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3} = \dfrac{\pi}{4}$.

Step 1 — check the window: $xy = \frac{1}{6} < 1$, so the plain formula applies.

Step 2 — apply it: $$\tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}} = \tan^{-1}1 = \frac{\pi}{4} \qquad \blacksquare$$

Example 3 — The $\pi$-correction in action

Evaluate $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$.

Step 1 — combine the last two, noting $xy = 6 > 1$: $$\tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}\frac{2 + 3}{1 - 6} = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$

Step 2 — add the first term: $\frac{\pi}{4} + \frac{3\pi}{4} = \pi$.

Answer: $\pi$ — a beautiful closed value, and unreachable without the correction term. (Companion fact: since the sum is $\pi$, the relation $x + y + z = xyz$ holds: $1 + 2 + 3 = 1 \cdot 2 \cdot 3$ ✓.)

Example 4 — Doubling a tangent inverse

Express $2\tan^{-1}\dfrac{1}{2}$ as a single inverse tangent.

Step 1 — the trio's tangent form ($\vert x \vert < 1$ ✓): $$2\tan^{-1}\frac{1}{2} = \tan^{-1}\frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \tan^{-1}\frac{1}{\frac{3}{4}} = \tan^{-1}\frac{4}{3}$$

Answer: $\tan^{-1}\dfrac{4}{3}$.

Example 5 — Recognising the trio's shapes

Evaluate $\sin^{-1}\dfrac{3}{5} - 2\tan^{-1}\dfrac{1}{3}$, using the fact that $\dfrac{3}{5} = \dfrac{2 \cdot \frac{1}{3}}{1 + \frac{1}{9}}$.

Step 1 — spot the shape: $\frac{3}{5} = \frac{2x}{1 + x^2}$ with $x = \frac{1}{3}$, and $\vert x \vert \leq 1$, so $$\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{1}{3}$$

Step 2 — subtract: the expression is $0$.

Answer: $0$ — shape-recognition converts what looks like a computation into an identity.

Example 6 — A subtraction identity

Prove that $\tan^{-1}\dfrac{2}{11} + \tan^{-1}\dfrac{7}{24} = \tan^{-1}\dfrac{1}{2}$.

Step 1 — window check: $xy = \frac{14}{264} < 1$ ✓.

Step 2 — apply the formula: $$\tan^{-1}\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11}\cdot\frac{7}{24}} = \tan^{-1}\frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} = \tan^{-1}\frac{125}{250} = \tan^{-1}\frac{1}{2} \qquad \blacksquare$$

Example 7 — The symmetric condition

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \dfrac{\pi}{2}$ (with $x, y, z > 0$), prove that $xy + yz + zx = 1$.

Step 1 — move one term across: $\tan^{-1}x + \tan^{-1}y = \frac{\pi}{2} - \tan^{-1}z = \cot^{-1}z = \tan^{-1}\frac{1}{z}$.

Step 2 — apply the addition formula and cross-multiply: $$\frac{x + y}{1 - xy} = \frac{1}{z} \implies z(x + y) = 1 - xy \implies xy + yz + zx = 1 \qquad \blacksquare$$

Answer: the condition every JEE "relation between $x, y, z$" question is fishing for. (Sanity check: $x = y = z = \frac{1}{\sqrt 3}$ gives sum $3 \cdot \frac{\pi}{6} = \frac{\pi}{2}$ and $3 \cdot \frac{1}{3} = 1$ ✓.)

Example 8 — An equation with root rejection

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

Step 1 — apply the addition formula (assuming the window, to be checked): $$\tan^{-1}\frac{5x}{1 - 6x^2} = \frac{\pi}{4} \implies \frac{5x}{1 - 6x^2} = 1$$

Step 2 — solve the quadratic: $6x^2 + 5x - 1 = 0$ factors as $(6x - 1)(x + 1) = 0$: candidates $x = \frac{1}{6}$, $\ x = -1$.

Step 3 — reject by substitution: $x = -1$ makes both terms negative — their sum cannot be $+\frac{\pi}{4}$ ✗. $\ x = \frac{1}{6}$: both terms positive and small, and the identity's window $6x^2 < 1$ holds ✓.

Answer: $x = \dfrac{1}{6}$ — the quadratic always manufactures a fake root in these equations; the rejection line is where the marks live.