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EAPCET Mathematics · Intermediate first year · Paper IA

Inverse Trigonometric Functions

Inverse Trigonometric Functions defines sin⁻¹, cos⁻¹, tan⁻¹ and the rest on restricted domains, and uses their properties to simplify expressions and solve equations.

Subject
Mathematics
Taught in
Intermediate first year · Paper IA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • Principal ranges: sin⁡−1x∈[−π2,π2]\sin^{-1}x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], cos⁡−1x∈[0,π]\cos^{-1}x \in [0, \pi], tan⁡−1x∈(−π2,π2)\tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)
  • sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2} for ∣x∣≤1|x| \le 1
  • tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2}
  • 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
  • sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x, while cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x
  • Principal ranges of the others: cot⁡−1x∈(0,π)\cot^{-1}x \in (0, \pi), sec⁡−1x∈[0,π]−{π2}\sec^{-1}x \in [0, \pi] - \left\{\tfrac{\pi}{2}\right\}, cosec⁡−1x∈[−π2,π2]−{0}\operatorname{cosec}^{-1}x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}
  • sec⁡−1x+cosec⁡−1x=π2\sec^{-1}x + \operatorname{cosec}^{-1}x = \tfrac{\pi}{2} for ∣x∣≥1|x| \ge 1
  • sin⁡−11x=cosec⁡−1x\sin^{-1}\tfrac{1}{x} = \operatorname{cosec}^{-1}x and cos⁡−11x=sec⁡−1x\cos^{-1}\tfrac{1}{x} = \sec^{-1}x for ∣x∣≥1|x| \ge 1; tan⁡−11x=cot⁡−1x\tan^{-1}\tfrac{1}{x} = \cot^{-1}x for x>0x > 0
  • 2tan⁡−1x=sin⁡−12x1+x22\tan^{-1}x = \sin^{-1}\dfrac{2x}{1 + x^{2}} for ∣x∣≤1|x| \le 1, =cos⁡−11−x21+x2= \cos^{-1}\dfrac{1 - x^{2}}{1 + x^{2}} for x≥0x \ge 0, =tan⁡−12x1−x2= \tan^{-1}\dfrac{2x}{1 - x^{2}} for ∣x∣<1|x| < 1
  • 2sin⁡−1x=sin⁡−1(2x1−x2)2\sin^{-1}x = \sin^{-1}\left(2x\sqrt{1 - x^{2}}\right) for ∣x∣≤12|x| \le \tfrac{1}{\sqrt{2}}; 2cos⁡−1x=cos⁡−1(2x2−1)2\cos^{-1}x = \cos^{-1}(2x^{2} - 1) for 0≤x≤10 \le x \le 1
  • sin⁡−1x+sin⁡−1y=sin⁡−1[x1−y2+y1−x2]\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left[x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}}\right] when xy≤0xy \le 0 or x2+y2≤1x^{2} + y^{2} \le 1; it is π−sin⁡−1[x1−y2+y1−x2]\pi - \sin^{-1}\left[x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}}\right] when x,y>0x, y > 0 and x2+y2>1x^{2} + y^{2} > 1
  • cos⁡−1x+cos⁡−1y=cos⁡−1[xy−1−x21−y2]\cos^{-1}x + \cos^{-1}y = \cos^{-1}\left[xy - \sqrt{1 - x^{2}}\sqrt{1 - y^{2}}\right] when x+y≥0x + y \ge 0
  • tan⁡−1x+tan⁡−1y+tan⁡−1z=tan⁡−1x+y+z−xyz1−xy−yz−zx\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\dfrac{x + y + z - xyz}{1 - xy - yz - zx} when the left side lies in (−π2,π2)\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)

What this chapter covers

  • Domains and principal value ranges
  • Properties and identities
  • Sum and difference formulas
  • Equations involving inverse trigonometric functions

How it is usually asked

  • Evaluate expressions such as sin(cos⁻¹(3/5))
  • Simplify sums of tan⁻¹ terms
  • Solve an equation in inverse trigonometric functions

Understanding Inverse Trigonometric Functions

The trigonometric functions are not one-one, so to invert them we restrict each to an interval where it is one-one and covers all its values. That interval is the principal value range: [−π/2, π/2] for sin⁻¹, [0, π] for cos⁻¹ and (−π/2, π/2) for tan⁻¹. So sin⁻¹x means the one angle in [−π/2, π/2] whose sine is x, not any angle with that sine.

Because of these ranges, sin(sin⁻¹x) = x for every x in [−1, 1], but sin⁻¹(sin θ) = θ only when θ is already in [−π/2, π/2]. Complementary results such as sin⁻¹x + cos⁻¹x = π/2 and the negative-argument rules come straight from the definitions. The sum formula tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] holds when xy < 1 and needs an extra π when x, y > 0 and xy > 1.

To evaluate an expression like cos(sin⁻¹(3/5)), name the angle, draw a right triangle with the given ratio and read off the other ratio, using the principal range to fix the sign. EAPCET questions are evaluations, simplifications of sums of tan⁻¹ terms and simple equations, and many wrong options are values that ignore the principal range.

Worked example

EAPCET-style question

The value of tan⁡−12+tan⁡−13\tan^{-1}2 + \tan^{-1}3 is

  1. (1)π4\dfrac{\pi}{4}
  2. (2)3π4\dfrac{3\pi}{4}
  3. (3)−π4-\dfrac{\pi}{4}
  4. (4)5π4\dfrac{5\pi}{4}

Solution

  1. Here x=2x = 2, y=3y = 3 and xy=6>1xy = 6 > 1 with x,y>0x, y > 0, so the sum formula needs π\pi added.
  2. x+y1−xy=51−6=−1\dfrac{x + y}{1 - xy} = \dfrac{5}{1 - 6} = -1, and tan⁡−1(−1)=−π4\tan^{-1}(-1) = -\tfrac{\pi}{4}.
  3. So the sum =π+(−π4)=3π4= \pi + \left(-\tfrac{\pi}{4}\right) = \tfrac{3\pi}{4}.
  4. Check: tan⁡−12\tan^{-1}2 and tan⁡−13\tan^{-1}3 both lie between π4\tfrac{\pi}{4} and π2\tfrac{\pi}{2}, so their sum lies between π2\tfrac{\pi}{2} and π\pi.

Answer: 3π4\dfrac{3\pi}{4}

Shortcuts and results to remember

  • sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x only for xx in [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]; on [π2,3π2]\left[\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right] it equals π−x\pi - x, so its graph is a zig-zag. Likewise cos⁡−1(cos⁡x)=2π−x\cos^{-1}(\cos x) = 2\pi - x on [π,2π][\pi, 2\pi] and tan⁡−1(tan⁡x)=x−π\tan^{-1}(\tan x) = x - \pi on (π2,3π2)\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right).
  • Telescoping: tan⁡−111+k(k+1)=tan⁡−1(k+1)−tan⁡−1k\tan^{-1}\dfrac{1}{1 + k(k + 1)} = \tan^{-1}(k + 1) - \tan^{-1}k, so such sums collapse to the last term minus the first.
  • (sin⁡−1x)2+(cos⁡−1x)2(\sin^{-1}x)^{2} + (\cos^{-1}x)^{2} lies in [π28,5π24]\left[\tfrac{\pi^{2}}{8}, \tfrac{5\pi^{2}}{4}\right] and (sin⁡−1x)3+(cos⁡−1x)3(\sin^{-1}x)^{3} + (\cos^{-1}x)^{3} lies in [π332,7π38]\left[\tfrac{\pi^{3}}{32}, \tfrac{7\pi^{3}}{8}\right].
  • For sin⁡−1\sin^{-1}(any expression), set the expression equal to a known constant α\alpha and solve, keeping α\alpha inside the allowed range.

More practice questions

Practice question 1

The value of sin⁡−1(sin⁡5)\sin^{-1}(\sin 5), with 5 in radians, is

  1. (1)5
  2. (2)π−5\pi - 5
  3. (3)5−2π5 - 2\pi
  4. (4)2π−52\pi - 5

Solution

  1. The answer must lie in [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]
  2. 5 lies between 3π2≈4.71\tfrac{3\pi}{2} \approx 4.71 and 5π2≈7.85\tfrac{5\pi}{2} \approx 7.85
  3. On this interval sin⁡−1(sin⁡x)=x−2π\sin^{-1}(\sin x) = x - 2\pi
  4. 5−2π≈−1.285 - 2\pi \approx -1.28, which is in [−π2,π2]\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], so the answer is 5−2π5 - 2\pi

Answer: 5−2π5 - 2\pi

Practice question 2

The sum to infinity of tan⁡−113+tan⁡−117+tan⁡−1113+…\tan^{-1}\tfrac{1}{3} + \tan^{-1}\tfrac{1}{7} + \tan^{-1}\tfrac{1}{13} + \dots is

  1. (1)π4\dfrac{\pi}{4}
  2. (2)π2\dfrac{\pi}{2}
  3. (3)π3\dfrac{\pi}{3}
  4. (4)π6\dfrac{\pi}{6}

Solution

  1. The kkth term is tan⁡−111+k(k+1)\tan^{-1}\dfrac{1}{1 + k(k + 1)}
  2. Write it as tan⁡−1(k+1)−k1+k(k+1)=tan⁡−1(k+1)−tan⁡−1k\tan^{-1}\dfrac{(k + 1) - k}{1 + k(k + 1)} = \tan^{-1}(k + 1) - \tan^{-1}k
  3. Sum of nn terms =tan⁡−1(n+1)−tan⁡−11= \tan^{-1}(n + 1) - \tan^{-1}1
  4. As n→∞n \to \infty this tends to π2−π4=π4\tfrac{\pi}{2} - \tfrac{\pi}{4} = \tfrac{\pi}{4}

Answer: π4\dfrac{\pi}{4}

Common mistakes to avoid

  • Applying tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] without checking xy < 1; when x, y > 0 and xy > 1, add π.
  • Writing sin⁻¹(sin 2π/3) = 2π/3; the answer must lie in [−π/2, π/2], so it is π/3.
  • Using cos⁻¹(−x) = −cos⁻¹x; the correct rule is cos⁻¹(−x) = π − cos⁻¹x, since cos⁻¹ values are never negative.
  • Reading sin⁻¹x as 1/sin x; the inverse function and the reciprocal (cosec x) are different things.

How to prepare it

To evaluate something like sin(cos⁻¹ x), draw a right triangle with the given ratio — it avoids sign mistakes.

Always check the answer lies in the principal range; that is where most options are designed to trap you.

Once the formulas feel familiar, test them the way the exam does: practise Inverse Trigonometric Functions questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Inverse Trigonometric Functions chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Inverse Trigonometric Functions part of the EAPCET Mathematics syllabus?
Yes. Inverse Trigonometric Functions is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Inverse Trigonometric Functions for EAPCET?
To evaluate something like sin(cos⁻¹ x), draw a right triangle with the given ratio — it avoids sign mistakes. Always check the answer lies in the principal range; that is where most options are designed to trap you.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Inverse Trigonometric Functions?
Applying tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] without checking xy < 1; when x, y > 0 and xy > 1, add π. Writing sin⁻¹(sin 2π/3) = 2π/3; the answer must lie in [−π/2, π/2], so it is π/3. Using cos⁻¹(−x) = −cos⁻¹x; the correct rule is cos⁻¹(−x) = π − cos⁻¹x, since cos⁻¹ values are never negative. Reading sin⁻¹x as 1/sin x; the inverse function and the reciprocal (cosec x) are different things.