Q.Solve the equation cos−1x+cos−12x=32π. OR Solve the equation sec−1(ax)−sec−1(bx)=sec−1b−sec−1a.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Inverse Cosine Addition
Inverse Cosine Addition – From Intuition to Formula
Suppose you know cosA=x and cosB=y and want the angle A+B — that is, cos−1x+cos−1y in terms of x and y.
The answer is not simply cos−1(xy−1−x21−y2) — that's the cosine of the sum, not the sum itself. The real formula is subtler, because inverse cosine returns an angle in a fixed range.
The Intuition
cos−1x is "the angle whose cosine is x", and by definition it lies in [0,π]. So cos−1x+cos−1y is a sum of two angles each in [0,π] — anywhere from 0 to 2π.
Inverse cosine is not linear, so take the cosine of the sum using the addition formula:
cos(cos−1x+cos−1y)=cos(cos−1x)cos(cos−1y)−sin(cos−1x)sin(cos−1y)
Since cos(cos−1x)=x and sin(cos−1x)=1−x2 (positive because cos−1x∈[0,π]):
cos(cos−1x+cos−1y)=xy−1−x21−y2
The sum itself is the inverse cosine of that expression — only if the sum lies in [0,π], the range of cos−1.
The Precise Statement
cos−1x+cos−1y=⎩⎨⎧cos−1(xy−1−x21−y2),2π−cos−1(xy−1−x21−y2),if x+y≥0if x+y<0
Why the case split? cos−1 always returns an angle in [0,π]. When x+y≥0 the sum lies in [0,π], so it equals the inverse cosine directly. When x+y<0 the sum lies in (π,2π), so we use cos−1(−t)=π−cos−1t to bring it back into range.
A common mistake is writing cos−1x+cos−1y=cos−1(xy−1−x21−y2) without checking x+y≥0. This is false when x+y<0 — you then need 2π minus that inverse cosine.
A Quick Example
Let x=y=−21. Then cos−1(−21)=32π, so the true sum is 34π. …
Setting A=cos−1x and B=cos−12x and taking the cosine of both sides of A+B=2π/3 turns the equation into an algebraic one solvable for x (the OR alternative equation is a separate problem, n …
Let A=cos−1x, B=cos−12x with A+B=32π; expand cos(A+B) and solve for x.
(Answering the primary equation; the OR alternative equation is a separate problem and is not required.)
Let A=cos−1x and B=cos−12x, so A+B=32π, hence cosA=x, cosB=2x, sinA=1−x2, sinB=1−4x2.
Taking cosine of both sides:
cos(A+B)=cos32π=−21
cosAcosB−sinAsinB=−21
2x2−(1−x2)(1−4x2)=−21
2x2+21=(1−x2)(1−4x2)
Squaring both sides:
4x4+2x2+41=(1−x2)(1−4x2)=1−5x2+4x4
2x2+41=1−5x2
7x2=43
x2=283
…
Showing the 12 most recent of 14 on this concept.
- CBSE 2026Set A1 markMCQQ.cos−1(cos67π)=(a) 6π(b) 3π(c) 65π(d) 67π
›Reveal solutionSolution
cos−1(cos67π)=65π.
The range of cos−1 is [0,π], but 67π∈/[0,π], so we cannot just cancel.
First find the actual cosine value:
cos67π=−cos6π=−23.
…
- CBSE 2025Set E1 markMCQQ.cos−1(cos58π)=(a) 58π(b) 52π(c) 5π(d) 53π
›Reveal solutionSolution
Reduce the angle to the principal range [0,π] of cos−1; the answer is 52π.
The angle 58π is not in [0,π], so we cannot just cancel. Use cos(2π−θ)=cosθ:
cos58π=cos(2π−58π)=cos52π.
…
- CBSE 2025Set E1 markMCQQ.sin{sin−151+cos−1x}=1, ⇒x=(a) 1(b) 0(c) 54(d) 51
›Reveal solutionSolution
Set the bracket equal to 2π and solve; x=51.
Since sinθ=1 only when θ=2π,
sin−151+cos−1x=2π.
…
- CBSE 2025Set E1 markMCQQ.cos−1x+sec−1x1=(a) 2π(b) cos−1(2x2−1)(c) cos−1(1−2x2)(d) cos−12x
›Reveal solutionSolution
sec−1x1=cos−1x, so the sum is 2cos−1x, which equals cos−1(2x2−1).
Use sec−1(y)=cos−1y1, so
sec−1x1=cos−1x.
Therefore
cos−1x+sec−1x1=2cos−1x. …
- CBSE 2024Set D1 markMCQQ.cos−1(−21)=(a) 32π(b) 3π(c) 6π(d) 2π
›Reveal solutionSolution
The principal value lies in [0,π]: cos−1(−21)=32π.
The range of cos−1 is [0,π]. Since cos32π=−21 and 32π∈[0,π]:
…
- CBSE 2024Set D1 markMCQQ.x∈[−1,1], cos−1x=(a) 2π−cot−1x(b) 2π−sin−1x(c) 2π−tan−1x(d) 2π−sec−1x
›Reveal solutionSolution
Complementary identity: cos−1x=2π−sin−1x for x∈[−1,1].
For x∈[−1,1] we have sin−1x+cos−1x=2π.
…
- CBSE 2024Set D1 markMCQQ.cos−1(cos67π)=(a) 67π(b) 65π(c) 3π(d) 6π
›Reveal solutionSolution
cos−1 must return a value in [0,π], so the answer is 65π, not 67π.
Since 67π∈/[0,π], we cannot simply cancel. Compute cos67π=−23.
…
- CBSE 2023Set E1 markMCQQ.x∈[−1,1], sin[2(sin−1x+cos−1x)]=(a) 0(b) 1(c) −1(d) 21
›Reveal solutionSolution
sin[2(sin−1x+cos−1x)]=0.
For all x∈[−1,1], sin−1x+cos−1x=2π. Hence
…
- CBSE 2023Set A1 markQ.Find the value of cos(sec−1x+cosec−1x), ∣x∣≥1.
›Reveal solutionSolution
sec−1x+cosec−1x=2π for ∣x∣≥1, so cos(sec−1x+cosec−1x)=cos2π=0.
…
- CBSE 2022Set HE2191 markQ.Fill in the blank: sin−1x+cos−1x= ______.
›Reveal solutionSolution
This is a standard inverse trigonometric identity valid for x∈[−1,1].
Let sin−1x=θ, so x=sinθ=cos(2π−θ), giving cos−1x=2π−θ. Adding, …
- CBSE 2021Set I1 markMCQQ.cos−1(2x)+sin−1(2x)=……, 2x∈[−1,1](a) 2π(b) 4π(c) π(d) 0
›Reveal solutionSolution
For all valid arguments, sin−1t+cos−1t=2π.
There is a standard identity: for every t∈[−1,1],
sin−1t+cos−1t=2π.
…
- CBSE 2019Set 65/3/11 markQ.If y=sin−1x+cos−1x, find dxdy.
›Reveal solutionSolution
The sum sin−1x+cos−1x is constant (π/2) for all x in [−1,1], so its derivative is zero: dxdy=0.
Concept and Intuition
The problem asks for the derivative of y=sin−1x+cos−1x. A brute-force approach would differentiate each inverse trig function separately using known formulas, then add. But that misses the deeper point.
There is a beautiful identity: for any x in [−1,1],
sin−1x+cos−1x=2π.
Why? Think geometrically. If θ=sin−1x, then sinθ=x and θ∈[−π/2,π/2]. The complementary angle π/2−θ has cosine equal to x, so cos−1x=π/2−θ. Adding them gives π/2.
Since y is constant, its derivative is zero — no calculation needed. This is the elegant, concept-first way.
Watch outA common mistake is to differentiate each term separately and get 1−x21−1−x21=0, which is correct but misses why the sum is constant. The identity is the real insight.
Step-by-Step Solution
- Recall the fundamental identity For any x∈[−1,1],
sin−1x+cos−1x=2π.
This holds because if sin−1x=θ, then cos(π/2−θ)=sinθ=x, and π/2−θ lies in [0,π], the principal range of cos−1. …
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