Q.Prove the following: sin−153+cos−11312=sin−16556
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🔒 Start your 14-day free trial to unlock the full solution →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). …
Convert both terms to angles with known sine/cosine components and add. …
Let sin−153=θ (so sinθ=53,cosθ=54) and cos−11312=ϕ (so cosϕ=1312,sinϕ=135), both in (0,2π). sin(θ+ϕ)=sinθcosϕ+cosθsinϕ=53⋅1312+54⋅135=6536+6520=6556. Since 0<θ+ϕ<π and this sum sits within the sine prin …
Name each angle, find its full sine/cosine pair using a Pythagorean triple, t …
Using the cosine-addition formula instead of the sine one, which is needed since the fin …
- 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). …
- 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. …
- 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). …
- 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. …
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