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Inverse Trigonometric Functions Set-1

March 26, 2026

Inverse Trigonometric Functions Set-1 📘

Did you know? Inverse trigonometric functions are the secret link between trigonometry and calculus. They appear directly in CBSE Class 12 boards, and indirectly in JEE & NEET questions involving limits, integrals, and coordinate geometry. Mastering just a few core identities can turn long questions into quick 2–3 line solutions.

1. Why Inverse Trigonometric Functions Matter 🎯

Inverse trigonometric functions help you “reverse” the usual trigonometric functions.

  • Trigonometric function: angle → ratio
  • Inverse trigonometric function: ratio → angle

You meet them in:

  • CBSE Class 12: A full chapter with definition, principal values, identities, and simple equations.
  • JEE Main/Advanced: In limits, integration, solution of equations, and graph-based questions.
  • NEET: Sometimes in physics (especially SHM, waves, and optics where arcsin/arctan forms appear).

So, this Set-1 is all about:

  • Understanding principal value branches
  • Learning standard values
  • Using basic identities to simplify expressions

2. Getting the Basics Clear 🧱

2.1 What is an inverse trigonometric function?

Example: Sine function

  • If sin θ = 1/2, then one of the values of θ is 30°.
  • But there are infinitely many such angles (…, -330°, -30°, 30°, 150°, 390°, …).

To define a single-valued inverse, we restrict the domain of the trigonometric function so that it becomes one-one and onto, and then define its inverse on that domain.

For sine:

  • Restricted domain: [π/2, π/2][- \pi/2,\ \pi/2]
  • Range of sin on this interval: [1, 1][-1,\ 1]

So the inverse sine function, written as sin⁻¹x or arcsin x, is defined from [1,1][-1, 1] to [π/2,π/2][- \pi/2, \pi/2].

In symbols:

sin1:[1,1][π2,π2]\sin^{-1} : [-1,1] \rightarrow \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Similarly, for other functions:

  • cos⁻¹x, tan⁻¹x, cot⁻¹x, sec⁻¹x, cosec⁻¹x

Each has its own domain and principal value range.


3. Principal Value Branches – Super Important for Exams 🔑

Memorising the principal value ranges is non‑negotiable. Many CBSE, JEE, and NEET maths questions are traps based on wrong range selection.

3.1 Quick Reference Table 📋

FunctionDomainPrincipal Value Range
sin⁻¹x-1 ≤ x ≤ 1π/2-\pi/2 to π/2\pi/2 (both included)
cos⁻¹x-1 ≤ x ≤ 10 to π\pi (both included)
tan⁻¹xall real xπ/2, π/2-\pi/2,\ \pi/2 (not including ends)
cot⁻¹xall real x0 to π\pi (not including ends)
sec⁻¹xx ≤ -1 or x ≥ 1[0, π]{π/2}[0,\ \pi] \setminus \{\pi/2\}
cosec⁻¹xx ≤ -1 or x ≥ 1[π/2, π/2]{0}[- \pi/2,\ \pi/2] \setminus \{0\}

Memory aid:

  • sin⁻¹ and tan⁻¹ → range around 0 symmetric: π/2-\pi/2 to π/2\pi/2
  • cos⁻¹ and cot⁻¹ → 0 to π\pi

4. Standard Values You Must Know 🧠

These are like tables for trigonometry, but now for inverse trig functions.

4.1 Basic angles table

xsin⁻¹xcos⁻¹xtan⁻¹x
00π/2\pi/20
1/2π/6\pi/6π/3\pi/3
√3/2π/3\pi/3π/6\pi/6
-1/2π/6-\pi/65π/65\pi/6
√3/3π/6\pi/6
1π/2\pi/20π/4\pi/4
-1π/2-\pi/2π\piπ/4-\pi/4

(“–” means that pair is not usually used in basic tables.)


5. Fundamental Identities – Your Shortcut Toolbox 🧰

These are “must‑remember” identities that help you simplify expressions.

5.1 Direct inverse identities

For x in the domain of the function:

sin(sin1x)=x\sin(\sin^{-1}x) = x cos(cos1x)=x\cos(\cos^{-1}x) = x tan(tan1x)=x\tan(\tan^{-1}x) = x

And for y in the principal value range:

sin1(siny)=y\sin^{-1}(\sin y) = y cos1(cosy)=y\cos^{-1}(\cos y) = y tan1(tany)=y\tan^{-1}(\tan y) = y

Note the condition:
The second set only works when y is within the principal range of the respective inverse function.


6. Classic Solved Examples (Step‑by‑Step) ✍️

Example 1: Evaluating a simple inverse trig value

Evaluate sin⁻¹(√3/2).

Step 1: Recall standard sine values
We know sin 60° = √3/2, i.e., sin(π/3) = √3/2.

Step 2: Check if angle is in principal range
Principal value range for sin⁻¹x is [π/2,π/2][- \pi/2, \pi/2].
π/3 lies in this interval.

Therefore,

sin1(32)=π3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}

Example 2: Carefully applying principal value

Evaluate cos⁻¹(-1/2).

Step 1: Know which angle gives cos θ = -1/2
cos 120° = -1/2, i.e., cos(2π/3) = -1/2.

Step 2: Check principal value range
For cos⁻¹x, range is [0, π].
2π/3 lies in this range.

So,

cos1(12)=2π3\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

Note: Even though cos 4π/3 = -1/2 also, we never use 4π/3 here because it lies outside [0, π].


Example 3: Composition involving tan and tan⁻¹

Simplify tan(tan⁻¹ 5).

Step 1: Identify the pattern
This is of the form tan(tan⁻¹ x).

For all real x:

tan(tan1x)=x\tan(\tan^{-1}x) = x

So,

tan(tan15)=5\tan(\tan^{-1}5) = 5

No need for detailed calculation.


Example 4: Composition in reverse order – more subtle

Simplify tan⁻¹(tan 4π/3).

Step 1: Identify principal value range
Range of tan⁻¹x is π/2, π/2-\pi/2,\ \pi/2.

Step 2: Reduce 4π/3 to this interval

4π/3 is in the 3rd quadrant.

We know tan(θ) has period π.
So subtract π:

4π/3π=π/34\pi/3 - \pi = \pi/3

And tan(4π/3) = tan(π/3).

Now π/3 is about 1.047, which is not in π/2, π/2-\pi/2,\ \pi/2 (since π/2 ≈ 1.57, actually π/3 is in this interval – so let’s reason directly).

Better route:

π/3 is indeed between -π/2 and π/2, so

tan1(tanπ/3)=π/3\tan^{-1}(\tan \pi/3) = \pi/3

But tan(4π/3) = tan(π/3), so:

tan1(tan4π/3)=tan1(tanπ/3)=π/3\tan^{-1}(\tan 4\pi/3) = \tan^{-1}(\tan \pi/3) = \pi/3

Hence,

tan1(tan4π/3)=π3\tan^{-1}(\tan 4\pi/3) = \frac{\pi}{3}

Key learning:
For tan⁻¹(tan θ), always convert θ to an equivalent angle in π/2, π/2-\pi/2,\ \pi/2.


7. Special Identities Connecting Different Inverses 🔄

Some very popular identities often used in JEE/boards:

  1. Relation between sin⁻¹ and cos⁻¹
sin1x+cos1x=π2,1x1\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad -1 \le x \le 1
  1. Relation between tan⁻¹ and cot⁻¹
tan1x+cot1x=π2,x>0\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x > 0
  1. Relation between tan⁻¹ and cot⁻¹ (alternate form)
tan1x+tan1(1x)=π2,x>0\tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}, \quad x > 0

and

tan1x+tan1(1x)=π2,x<0\tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2}, \quad x < 0

Quick Revision Box 📦

  • sin⁻¹x + cos⁻¹x = π/2
  • tan⁻¹x + cot⁻¹x = π/2
  • Always think: “Is my final angle in the principal range?”
  • For sin⁻¹x: angle in [π/2,π/2][- \pi/2, \pi/2]
  • For cos⁻¹x: angle in [0, π]
  • For tan⁻¹x: angle in π/2,π/2-\pi/2, \pi/2

8. Mixed Example Using Identities 🧮

Example 5: Evaluate sin⁻¹(3/5) + cos⁻¹(3/5)

We directly use the identity:

sin1x+cos1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

So, taking x = 3/5:

sin1(35)+cos1(35)=π2\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \frac{\pi}{2}

No calculation required.


Example 6: Simplify tan⁻¹(1/3) + tan⁻¹(2/5)

We can use the formula:

tan1a+tan1b=tan1(a+b1ab)\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)

provided that the product ab < 1 and the sum lies in the principal range.

Here, a = 1/3, b = 2/5.

Compute:

  • a + b = 1/3 + 2/5 = (5 + 6)/15 = 11/15
  • ab = (1/3) × (2/5) = 2/15 < 1, so formula is valid.

So,

tan1(13)+tan1(25)=tan1(11/1512/15)\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{2}{5}\right) = \tan^{-1}\left(\frac{11/15}{1 - 2/15}\right)

Now,

1 - 2/15 = 13/15

Thus:

11/1513/15=1113\frac{11/15}{13/15} = \frac{11}{13}

Therefore,

tan1(13)+tan1(25)=tan1(1113)\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{2}{5}\right) = \tan^{-1}\left(\frac{11}{13}\right)

Since 11/13 is positive and tan⁻¹x of a positive number lies in (0, π/2), which is in principal range, this is our final answer.


9. Typical Mistakes Students Make (and How to Avoid Them) 🚫

Mistake TypeWhat Students Do WrongHow to Fix It
Ignoring principal rangeWrite sin⁻¹(√3/2) = 5π/3Always choose angle inside range of sin⁻¹, cos⁻¹, tan⁻¹
Forgetting identitiesTry to calculate each part separatelyMemorise core identities like sin⁻¹x + cos⁻¹x = π/2
Wrong sign in tan⁻¹ identitiesUse same formula for x > 0 and x < 0Remember sign change for x < 0 in tan⁻¹x + tan⁻¹(1/x)
Taking multiple values of angleGive all general solutions (like 2nπ + …) in inverse questionsIn inverse trig, we only write principal value, not general
Mixing degrees and radiansSubstitute π/6 as 30 (or vice versa) carelesslyStick to radians for Class 12, JEE, and NEET maths

10. Strategy Tips for CBSE, JEE & NEET 🧠⚙️

For CBSE Class 12

  • Focus on:
    • Definitions, domains, ranges
    • Simple identities and compositions
    • Proving small identities using basic trigonometry
  • Practice NCERT Examples + Exercise questions thoroughly.
  • In board exams, inverse trig questions are usually straightforward: evaluate, simplify, or prove.

For JEE Main / Advanced

  • Expect:
    • Expressions like tan⁻¹x + tan⁻¹y
    • Transformations using identities
    • Links with algebra (substitutions like x = tan θ)
  • Always think:
    • Can I convert this expression into a single inverse trig term?
    • Is there a smart substitution using right triangle geometry?

For NEET Aspirants

  • Direct inverse trig is rare in NEET maths, but:
    • It shows up in physics, especially when solving for angles from ratios.
    • Being comfortable with principal values helps avoid conceptual errors.

11. Visualising Inverse Trigonometric Functions (In Your Mind) 🧠🎨

You don’t need to draw perfect graphs in the exam, but a mental picture helps.

sin⁻¹x graph

  • Domain: -1 to 1
  • Range: -π/2 to π/2
  • Increasing curve passing through origin (0, 0).

Imagine taking the graph of y = sin x on [π/2, π/2][- \pi/2,\ \pi/2] and reflecting it in the line y = x. That reflection is y = sin⁻¹x.

tan⁻¹x graph

  • Domain: all real numbers
  • Range: -π/2 to π/2
  • Passes through (0, 0)
  • Has horizontal asymptotes at y = π/2 and y = -π/2

This helps you guess the sign and approximate size of angles when x is large or small.


12. Mini Practice Drill 💪

Try these without looking at the solutions first:

  1. Evaluate cos⁻¹(√3/2).
  2. Find the value of sin⁻¹(1) + cos⁻¹(1).
  3. Simplify tan⁻¹(2) + tan⁻¹(3).
  4. Find the principal value of sin⁻¹(sin 7π/6).
  5. If sin⁻¹x + cos⁻¹x = α, what is α?

Quick Answers (for self-check)

  1. cos⁻¹(√3/2) = π/6
  2. sin⁻¹(1) = π/2, cos⁻¹(1) = 0 → sum = π/2
  3. Use formula: tan⁻¹a + tan⁻¹b = tan⁻¹((a + b)/(1 - ab))
    • Here a = 2, b = 3 → (a + b) = 5, ab = 6
    • So 1 - ab = -5
    • Expression = tan⁻¹(5 / -5) = tan⁻¹(-1)
    • tan⁻¹(-1) = -π/4 (since -1 lies in principal range)
  4. sin 7π/6 = sin(π + π/6) = -1/2
    • sin⁻¹(-1/2) = -π/6 (principal value)
  5. Always α = π/2, since sin⁻¹x + cos⁻¹x = π/2

13. How to Tackle “Set‑1” Type Questions Efficiently ⏱️

When you see objective questions (like in a quiz or MCQ test) on this topic:

  1. Scan for pattern

    • Is it of type sin⁻¹x + cos⁻¹x?
    • Or tan⁻¹a + tan⁻¹b?
    • Or tan⁻¹(tan θ)?
  2. Immediately recall identity

    • Don’t re‑derive in the exam; directly apply.
  3. Check range at the end

    • If your angle is outside the principal range, bring it inside using periodicity.
  4. Use right triangle logic for simple ratios

    • For sin⁻¹(3/5), think of a right triangle with opposite = 3, hypotenuse = 5.

This habit will save precious minutes in both boards and entrance exams.


14. Ready to Test Yourself? 🚀

Now that you’ve revised the core ideas, identities, and typical questions from Inverse Trigonometric Functions Set‑1, it’s the perfect time to evaluate your understanding with a focused practice quiz.

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