Q.Find the value, if it exists. If not, give the reason for non-existence.
Concept understanding — Properties of Inverse Trigonometric Functions
These are the identity-level properties (Properties I–V of the chapter) that let an inverse-trig expression be simplified WITHOUT drawing a triangle or invoking a sum formula — they hold strictly within the principal value branches.
Property I — undoing the outer inverse. f−1(f(θ))=θ holds only when θ already lies in f's principal domain: sin−1(sinθ)=θ if θ∈[−2π,2π]; cos−1(cosθ)=θ if θ∈[0,π]; tan−1(tanθ)=θ if θ∈(−2π,2π); similarly for cosec−1,sec−1,cot−1 on their own principal domains. If θ is OUTSIDE the principal domain, f−1(f(θ))=θ — instead, use periodicity/symmetry to rewrite f(θ) as f(θ1) for some θ1 that IS inside the principal domain, then f−1(f(θ))=θ1. E.g. sin−1(sin65π)=sin−1(sin(π−6π))=sin−1(sin6π)=6π, since 6π∈[−2π,2π].
Property II — undoing the inner inverse. f(f−1(x))=x holds throughout f−1's entire domain with no extra condition: sin(sin−1x)=x for x∈[−1,1]; cos(cos−1x)=x for x∈[−1,1]; tan(tan−1x)=x for every real x; and likewise for the other three on their own domains. This is the direct definition of "inverse" and never needs a range check.
Property III (reciprocal identities). sin−1(x1)=cosec−1x and cos−1(x1)=sec−1x, both for x∈R∖(−1,1); and tan−1(x1)=cot−1x if x>0, but =−π+cot−1x if x<0 (the sign correction is needed here because tan−1(x1) stays in (−2π,0) for x<0 while cot−1x lands in (2π,π) — different branches of the same underlying angle).
Property IV (reflection identities — negating the argument). sin−1(−x)=−sin−1x; tan−1(−x)=−tan−1x; cosec−1(−x)=−cosec−1x (all three odd); but cos−1(−x)=π−cos−1x; sec−1(−x)=π−sec−1x; cot−1(−x)=π−cot−1x (all three pick up a π−, since their principal range [0,π]-type interval isn't symmetric about 0).
Property V (cofunction identities — complementary pairs sum to π/2). sin−1x+cos−1x=2π for x∈[−1,1]; tan−1x+cot−1x=2π for every real x; cosec−1x+sec−1x=2π for x∈R∖(−1,1). These three are the most-used single identity in the whole chapter — whenever a sum/difference of a function AND its cofunction appears with the same argument, it collapses immediately to ±2π without any further work.
Properties I and II look symmetric but are NOT interchangeable: Property II is unconditional, Property I demands a range check first. Mixing them up is the single most common inverse-trig error — e.g. wrongly writing cos−1(cos67π)=67π (Property I misapplied, since 67π∈/[0,π]) instead of correctly reducing cos67π=cos(2π−67π)=cos65π and answering 65π.
Search queries such as "inverse trigonometric functions properties class 12 formula" and "inverse trig important questions" are extremely common around board-exam and JEE Main preparation time, since these five identity properties form the core of the NCERT/CBSE Class 12 Inverse Trigonometric Functions chapter. The Property I versus Property II distinction highlighted here is one of the most frequently tested traps in both CBSE board papers and competitive-exam MCQs.
Simplify the inner trig value first, then check whether it lies in the outer inverse function's domain (all three do here).
- −2π.
- −4π.
- 5−2π.
Each part reduces the inner expression to a single number, and since that number always lands inside the relevant domain, all three values exist.
Step 1. (i) Simplify the inner value. cosπ=−1.
Step 2. (i) Apply sin−1. −1∈[−1,1] (the domain of sin−1), so sin−1(cosπ)=sin−1(−1)=−2π. It exists.
Step 3. (ii) Simplify the inner value. sin(−25π)=sin(−25π+2π)=sin(−2π)=−1.
Step 4. (ii) Apply tan−1. tan−1 accepts every real number, so tan−1(−1)=−4π. It exists.
Step 5. (iii) Reduce 5 radians modulo 2π toward the principal range. 5−2π≈5−6.2832=−1.2832 rad, and −2π≈−1.5708≤−1.2832≤2π≈1.5708, so 5−2π∈[−2π,2π].
Step 6. (iii) Conclude. Since sin(5−2π)=sin5 (periodicity) and 5−2π lies in the principal range, sin−1(sin5)=5−2π. It exists.
(i) −2π. (ii) −4π. (iii) 5−2π.
Reduce the inner trig value first, then check domain/range before applying the inverse function
- Assuming sin−1(sin5)=5 directly, without checking that 5 radians itself is far outside [−π/2,π/2]
- Forgetting to reduce −5π/2 by a full period 2π before evaluating the sine
- CBSE 2026Set ANNUAL1 markMCQQ.If x<0, then tan−1(x1) is equal to :(a) −π+cot−1(x)(b) tan−1(x)(c) −π+tan−1x(d) cot−1(x)
›Reveal solutionSolution
Derives the identity relating tan−1(1/x) and cot−1x for negative x by comparing ranges, then confirms with a numeric check.
- For x>0, the standard identity is tan−1(x1)=cot−1x, both lying in (0,2π).
- For x<0, x1<0 too, so tan−1(x1)∈(−2π,0) (using the principal range of tan−1).
- But cot−1(x) for x<0 lies in (2π,π) (principal range of cot−1 is (0,π)), so cot−1(x) itself cannot equal tan−1(1/x) directly — it must be shifted by π to land in the correct range: tan−1(x1)=cot−1(x)−π=−π+cot−1(x).
- Numeric check with x=−2: tan−1(−21)≈−0.4636 rad. Also cot−1(−2)=π−cot−1(2)≈3.1416−0.4636=2.6779, so −π+2.6779≈−0.4636 — matches exactly.
✓Final answer(a) −π+cot−1(x)
- CBSE 2024Set ANNUAL1 markMCQQ.If sin−1x+cot−1(21)=2π, then x is equal to :(a) 52(b) 21(c) 23(d) 51
›Reveal solutionSolution
Uses sin−1x+cos−1x=π/2 together with a right-triangle reading of cot−1(1/2) as cos−1 of something.
- Since sin−1x+cos−1x=2π for all x∈[−1,1], the given equation sin−1x+cot−1(21)=2π means cot−1(21)=cos−1x.
- Let θ=cot−1(21), so cotθ=21, i.e. tanθ=2. In a right triangle, opposite =2, adjacent =1, hypotenuse =1+4=5.
- So cosθ=51 (with θ∈(0,π), the range of cot−1, and cotθ>0 places θ in the first quadrant where cosine is positive).
- Since θ=cos−1x, we get x=cosθ=51.
✓Final answer(d) 51
- CBSE 2023Set ANNUAL1 markMCQQ.If 3cos−1x=cos−1(4x3−3x),(a) x∈(21,1)(b) x∈[21,1](c) x∈(−∞,1](d) x∈[21,∞)
›Reveal solutionSolution
The triple-angle identity cos3θ=4cos3θ−3cosθ only matches cos−1's principal branch when 3θ∈[0,π].
- Let x=cosθ with θ=cos−1x∈[0,π] (the principal branch of cos−1).
- The identity cos3θ=4cos3θ−3cosθ gives 4x3−3x=cos3θ.
- For the given equation 3cos−1x=cos−1(4x3−3x) to hold, we need cos−1(cos3θ)=3θ, which is true only when 3θ itself lies in [0,π] (the range of cos−1).
- 3θ∈[0,π]⟺θ∈[0,3π].
- Since θ=cos−1x is a decreasing function, θ∈[0,3π] corresponds to x=cosθ∈[cos3π,cos0]=[21,1].
✓Final answer(b) x∈[21,1]
- CBSE 2020Set ANNUAL1 markMCQQ.If sin−1x+sin−1y=32π, then cos−1x+cos−1y is equal to :(a) π(b) 32π(c) 3π(d) 6π
›Reveal solutionSolution
Using the identity sin−1x+cos−1x=π/2 applied to both x and y, cos−1x+cos−1y=π−32π=3π.
- Recall the standard identity: for any x∈[−1,1], sin−1x+cos−1x=2π.
- Apply this identity to x: sin−1x+cos−1x=2π.
- Apply the same identity to y: sin−1y+cos−1y=2π.
- Add these two equations: (sin−1x+sin−1y)+(cos−1x+cos−1y)=2π+2π=π.
- We are given sin−1x+sin−1y=32π. Substitute this in: 32π+(cos−1x+cos−1y)=π.
- Solve for the required sum: cos−1x+cos−1y=π−32π=33π−2π=3π.
✓Final answercos−1x+cos−1y=3π — option (c).
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