JEE Weightage:

JEE Maths — Inverse Trigonometry

JEE Maths — Inverse Trigonometry — JEE strategy, weightage, PYQs, traps

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Tags Inverse Trig Jee

Chapter Overview & Weightage

Inverse Trigonometry is a 1-question-guaranteed chapter in JEE Main and a frequent appearance in JEE Advanced. It blends with calculus (limits, integration, differentiation of inverse trig functions), so high competence here pays compound returns through the rest of the maths section.

JEE Main Weightage (Year-by-Year)

YearQuestionsMarksSub-topics
202414Domain/range manipulation
202328Sum-of-arctan, equation solving
202214Identity-based simplification
202114tan⁡−1+tan⁡−1\tan^{-1} + \tan^{-1}
202028Range of compound function, integral application

Key Concepts You Must Know

Principal value branches:

  • sin⁡−1:[−1,1]→[−π/2,π/2]\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2]
  • cos⁡−1:[−1,1]→[0,π]\cos^{-1}: [-1, 1] \to [0, \pi]
  • tan⁡−1:R→(−π/2,π/2)\tan^{-1}: \mathbb{R} \to (-\pi/2, \pi/2)
  • cot⁡−1:R→(0,π)\cot^{-1}: \mathbb{R} \to (0, \pi)
  • sec⁡−1:∣x∣≥1→[0,π]∖{π/2}\sec^{-1}: |x| \geq 1 \to [0, \pi] \setminus \{\pi/2\}
  • csc⁡−1:∣x∣≥1→[−π/2,π/2]∖{0}\csc^{-1}: |x| \geq 1 \to [-\pi/2, \pi/2] \setminus \{0\}

Sum and difference formulas:

tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy(if xy<1)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy} \quad (\text{if } xy < 1)

sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2)\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})

Identity: sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2 for all x∈[−1,1]x \in [-1, 1].

Negative arguments: sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x, tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x, but cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x.

Important Formulas

tan⁡−1x+tan⁡−1y={tan⁡−1x+y1−xyxy<1π+tan⁡−1x+y1−xyxy>1,x>0−π+tan⁡−1x+y1−xyxy>1,x<0\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} & xy < 1 \\ \pi + \tan^{-1}\dfrac{x+y}{1-xy} & xy > 1, x > 0 \\ -\pi + \tan^{-1}\dfrac{x+y}{1-xy} & xy > 1, x < 0 \end{cases}

The “if xy<1xy < 1” caveat is what trips students up. Always check the sign of xyxy before applying the formula.

2tan⁡−1x=tan⁡−12x1−x2(∣x∣<1)2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2} \quad (|x| < 1)

2tan⁡−1x=sin⁡−12x1+x2(∣x∣≤1)2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} \quad (|x| \leq 1)

2tan⁡−1x=cos⁡−11−x21+x2(x≥0)2\tan^{-1}x = \cos^{-1}\frac{1-x^2}{1+x^2} \quad (x \geq 0)

tan⁡−1x=sin⁡−1x1+x2=cos⁡−111+x2\tan^{-1}x = \sin^{-1}\frac{x}{\sqrt{1+x^2}} = \cos^{-1}\frac{1}{\sqrt{1+x^2}}

sin⁡−1x=tan⁡−1x1−x2\sin^{-1}x = \tan^{-1}\frac{x}{\sqrt{1-x^2}}

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 January Shift 2

Find tan⁡−112+tan⁡−113\tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3}.

Here xy=1/6<1xy = 1/6 < 1, so we can apply the simple sum formula:

tan⁡−11/2+1/31−1/6=tan⁡−15/65/6=tan⁡−1(1)=π4\tan^{-1}\frac{1/2 + 1/3}{1 - 1/6} = \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}(1) = \frac{\pi}{4}

PYQ 2 — JEE Main 2023 April Shift 1

Find the principal value of cos⁡−1(cos⁡(7π/6))\cos^{-1}(\cos(7\pi/6)).

7π/67\pi/6 is outside the principal range [0,π][0, \pi] for cos⁡−1\cos^{-1}. Reduce: cos⁡(7π/6)=cos⁡(π+π/6)=−cos⁡(π/6)=−3/2\cos(7\pi/6) = \cos(\pi + \pi/6) = -\cos(\pi/6) = -\sqrt{3}/2.

So cos⁡−1(cos⁡(7π/6))=cos⁡−1(−3/2)=π−π/6=5π/6\cos^{-1}(\cos(7\pi/6)) = \cos^{-1}(-\sqrt{3}/2) = \pi - \pi/6 = 5\pi/6.

PYQ 3 — JEE Advanced 2022

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

Use cos⁡−1x=π/2−sin⁡−1x\cos^{-1}x = \pi/2 - \sin^{-1}x:

sin⁡−1x+sin⁡−1(1−x)=π/2−sin⁡−1x\sin^{-1}x + \sin^{-1}(1-x) = \pi/2 - \sin^{-1}x

2sin⁡−1x+sin⁡−1(1−x)=π/22\sin^{-1}x + \sin^{-1}(1-x) = \pi/2

Let sin⁡−1x=α\sin^{-1}x = \alpha, then sin⁡−1(1−x)=π/2−2α\sin^{-1}(1-x) = \pi/2 - 2\alpha, so 1−x=cos⁡2α=1−2sin⁡2α=1−2x21 - x = \cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2x^2.

So x=2x2  ⟹  x(2x−1)=0  ⟹  x=0x = 2x^2 \implies x(2x - 1) = 0 \implies x = 0 or x=1/2x = 1/2.

Verify both: x=0x = 0 gives 0+π/2=π/20 + \pi/2 = \pi/2 ✓; x=1/2x = 1/2 gives π/6+π/6=π/3\pi/6 + \pi/6 = \pi/3, and RHS =cos⁡−1(1/2)=π/3= \cos^{-1}(1/2) = \pi/3 ✓. Both work.

Difficulty Distribution

Difficulty% of QuestionsSub-topics
Easy30%Direct sum/difference, principal values
Medium50%Identity-based simplification, conversion
Hard20%Equations involving multiple inverse trigs, JEE Advanced multi-step

Expert Strategy

Week 1 — Memorise principal value ranges and identities. Drill the differences between sin⁡−1(−x)\sin^{-1}(-x) and cos⁡−1(−x)\cos^{-1}(-x).

Week 2 — Sum/difference formulas and the “if xy < 1” caveat. Solve 30+ problems mixing the standard form and the corrected form.

Week 3 — Mixed problems combining inverse trig with calculus. Domain/range questions, equations, and definite integrals involving inverse trig.

Topper’s substitution trick: for tan⁡−12x1−x2\tan^{-1}\dfrac{2x}{1-x^2}, set x=tan⁡θx = \tan\theta and the expression becomes 2θ=2tan⁡−1x2\theta = 2\tan^{-1}x. Same trick works for sin⁡−12x1+x2\sin^{-1}\dfrac{2x}{1+x^2}. Saves enormous time on JEE Main.

Common Traps

Trap 1: Applying tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x+y}{1-xy} when xy>1xy > 1.

Without the ±π\pm \pi correction, the answer falls outside the principal range. Always check xyxy vs 1 before applying.

Trap 2: Treating sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for all xx.

This holds only for x∈[−π/2,π/2]x \in [-\pi/2, \pi/2]. Outside this range, you must reduce xx to the principal range first.

Trap 3: Wrong negative-argument rule for cos⁡−1\cos^{-1}.

cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x, not −cos⁡−1x-\cos^{-1}x. Different from sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1}.

Trap 4: Forgetting to verify solutions.

In equations like PYQ 3 above, always check that solutions lie in the domains of all inverse trig functions involved. Extraneous solutions arise from squaring or sin/cos manipulation.

Trap 5: Confusing sec⁡−1x\sec^{-1}x vs cos⁡−1(1/x)\cos^{-1}(1/x).

They differ by domain considerations. Be careful when converting between secant and cosine inverses, especially with negative arguments.