Q.Prove that sec−1x+cosec−1x=2π, ∀∣x∣≥1
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Inverse Function Relationship
Two functions are inverses when each undoes the other. If f sends a to b, then f−1 sends b back to a. Chain them together and you land exactly where you started.
The defining equations
If f−1 is the inverse of f, then
f−1(f(x))=xandf(f−1(y))=y.
The first holds for every x in the domain of f; the second for every y in the range of f. This "round trip returns the input" is what inverse really means.
When does an inverse exist?
Only a one-to-one function (distinct inputs give distinct outputs) can be inverted — otherwise some output would have to map back to two inputs, which no function allows. Graphically, f must pass the horizontal line test.
When a function is not one-to-one over its whole domain (like sinx or x2), we first restrict it to a piece where it is, and the inverse lives on that restricted piece.
The geometry
Because (a,b) lies on f exactly when (b,a) lies on f−1, the graph of f−1 is the mirror image of f across the line y=x. Consequently the domain and range swap: the range of f becomes the domain of f−1.
Why the restriction bites — the trig case
For inverse trigonometric functions the relationship is one-sided. The "outer undo" always works:
sin(sin−1x)=xfor all x∈[−1,1].
But the "inner undo" only works on the principal range:
sin−1(sinx)=xonly if x∈[−2π,2π]. …
Setting θ=sec−1x and using the co-function identity connecting cosine and sine of complementary angles shows directly that sec−1x and cosec−1x must sum to a fixed value. …
Set θ=sec−1x, express 1/x as a cosine, rewrite it as a sine of the complementary angle, and identify that angle as cosec−1x.
Let θ=sec−1x, so by definition secθ=x, where θ∈[0,π]∖{π/2}.
Then:
cosθ=x1
Using the co-function identity cosθ=sin(2π−θ):
sin(2π−θ)=x1
Taking reciprocal (cosecant):
cosec(2π−θ)=x
By definition of cosec−1:
2π−θ=cosec−1x
So:
θ+cosec−1x=2π
Substituting back θ=sec−1x: …
Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The value of \tan^{-1}\left(\frac{1}{x}\right) will be:(a) \tan x(b) \cot x(c) \cot^{-1}\frac{1}{x}(d) \cot^{-1} x
›Reveal solutionSolution
tan−1(1/x) and cot−1x are the same angle for positive x, because tangent and cotangent are reciprocal functions.
Concept: For x>0, the standard identity is
tan−1(x1)=cot−1x …
- CBSE 2025Set ANNUAL1 markMCQQ.Value of tan−1(tan67π) is:(a) 6π(b) 67π(c) 31(d) 65π
›Reveal solutionSolution
Use periodicity of tan to bring the angle into the principal range (−2π,2π) of tan−1.
Since tan has period π: tan67π=tan(67π−π)=tan6π.
…
- CBSE 2025Set A1 markQ.If −1≤x≤1, then sin(sin−1x)= ______.
›Reveal solutionSolution
sin and sin−1 are inverse functions of each other on [−1,1], so applying one after the other returns the original input.
For −1≤x≤1, sin−1x is defined and gives an angle θ∈[−2π,2π] such that sinθ=x. Applying sin to this angle simply recovers x: …
- CBSE 2025Set ANNUAL1 markMCQQ.The principal value of sin−1(sin32π)+tan−1(tan43π) is(a) 3π(b) 12π(c) 125π(d) 127π
›Reveal solutionSolution
Bring each angle into the principal-value range using sin−1(sinθ)=π−θ and tan−1(tanθ)=θ−π for θ in the second quadrant.
Term 1: sin−1(sin32π)
The range of sin−1 is [−2π,2π]. Since 32π∈(2π,π) lies outside this range, use
sin−1(sinθ)=π−θfor θ∈(2π,π)
sin−1(sin32π)=π−32π=3π
Term 2: tan−1(tan43π)
…
- CBSE 2024Set D1 markMCQQ.x∈[−1,1], sin−1(−x)=(a) −sin−1x(b) sin−1x(c) −cos−1x(d) cos−1x
›Reveal solutionSolution
sin−1 is odd: sin−1(−x)=−sin−1x.
For x∈[−1,1], the inverse sine is an odd function, so
…
- CBSE 2024Set ANNUAL1 markQ.The value of sin−1(sin32π) is ________.
›Reveal solutionSolution
Since 32π is outside the principal range of sin−1, first rewrite sin32π as the sine of an angle inside [−2π,2π].
sin32π=sin(π−32π)=sin3π
…
- CBSE 2024Set ANNUAL1 markMCQQ.The principal value of tan−1(tan67π)(a) 6π(b) 4π(c) 3π(d) 2π
›Reveal solutionSolution
Reduce the angle into the principal branch (−2π,2π) using the period π of tan.
tan−1(tanθ)=θ holds only when θ already lies in the principal value branch (−2π,2π).
Here θ=67π, which does not lie in (−2π,2π). Since tan has period π:
tan67π=tan(67π−π)=tan6π …
- CBSE 2023Set E1 markMCQQ.sin(sin−121)=(a) 1(b) 21(c) 23(d) 0
›Reveal solutionSolution
sin(sin−121)=21.
Since 21∈[−1,1], sin(sin−1x)=x holds:
…
- CBSE 2022Set ANNUAL1 markMCQQ.Write the value of sin−131+cos−131.(a) 0(b) 1(c) 2π(d) π
›Reveal solutionSolution
Use the standard identity sin−1x+cos−1x=2π for any x∈[−1,1].
Here x=31∈[−1,1], so directly:
…
- CBSE 2022Set ANNUAL1 markMCQQ.If sin−1x=5π, then what is the value of cos−1x?(a) 109π(b) 107π(c) 105π(d) 103π
›Reveal solutionSolution
Use sin−1x+cos−1x=2π to get cos−1x directly from sin−1x.
Given sin−1x=5π.
…
- CBSE 2022Set ANNUAL1 markQ.If sin−15x+sec−145=2π, then x= ____. Choices given: [1, 3, 4, 5]
›Reveal solutionSolution
Rewrite sec−1 in terms of cos−1 and use sin−1z+cos−1z=2π.
sec−145=cos−154 (since sec−1y=cos−1y1).
Given: sin−15x+cos−154=2π.
…
- CBSE 2022Set ANNUAL1 markMCQQ.cos−1(cos67π) is equal to OR If f:R→R be given by f(x)=(3−x3)1/3, then f∘f(x) is(a) x1/3(b) x3(c) x(d) None of the above(a) 67π(b) 65π(c) 3π(d) 6π
›Reveal solutionSolution
cos−1(cosθ) returns the angle in the principal range [0,π] having the same cosine; 67π lies outside it.
The principal-value range of cos−1 is [0,π], but 67π∈/[0,π]. Compute the cosine:
cos67π=cos(π+6π)=−cos6π=−23.
We need the angle θ∈[0,π] with cosθ=−23. That angle is
θ=π−6π=65π.
Hence cos−1(cos67π)=65π, option (b).
…
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