Q.(i) Solve tan−1(2x)+tan−1(3x)=4π.
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Inverse Tangent Identities
The inverse tangent function obeys a family of addition and doubling identities that let you combine two arctangents into one. They come straight from the tangent addition formula, but they carry conditions you must respect.
The core addition identity
Start from tan(A+B)=1−tanAtanBtanA+tanB. Put A=tan−1x and B=tan−1y, so tanA=x and tanB=y. Then
tan−1x+tan−1y=tan−1(1−xyx+y),xy<1.
The restriction xy<1 keeps the combined angle inside the principal range (−π/2,π/2).
If xy>1 the raw formula lands in the wrong branch, so you must correct it:
tan−1x+tan−1y=π+tan−1(1−xyx+y) (x,y>0),
and −π+tan−1(⋅) when x,y<0. Ignoring this is the classic exam slip.
Subtraction
Replacing y with −y gives
tan−1x−tan−1y=tan−1(1+xyx−y),xy>−1.
The doubling identity
Set y=x in the addition formula:
2tan−1x=tan−1(1−x22x),−1<x<1.
The same angle can also be rewritten through sine and cosine, which is handy in integration and in proofs -- but each alternate form only matches 2tan−1x on part of its domain, so the two forms carry different conditions:
2tan−1x=sin−1(1+x22x),−1≤x≤1,
2tan−1x=cos−1(1+x21−x2),x≥0.
The cos−1 form needs x≥0 -- it fails for negative x. Check x=−1: 2tan−1(−1)=2(−4π)=−2π, but cos−1(1+11−1)=cos−1(0)=2π, the wrong sign entirely. The sin−1 form has no such restriction because sin−1 (unlike cos−1) can return a negative angle.
The complementary identity
For every real x,
tan−1x+cot−1x=2π.
This holds without restriction because tan−1 and cot−1 of the same value are complementary angles. …
(i) tan−12x+tan−13x=4π gives 6x2+5x−1=0, so x=61 (reject x=−1). (ii) For f(x)=x2−4x−3 on [1,4], 3f(4)−f(1)=1=f′(c)=2c−4⇒c=25∈(1,4), v …
(i) Combine the inverse tangents and solve the quadratic, keeping the valid root x=61. (ii) Compute the average slope =1 and find c=25 where f′(c)=1, confirming the Mean Value Theorem.
Part (i). Using tan−1A+tan−1B=tan−11−ABA+B:
1−6x22x+3x=tan4π=1 ⇒ 5x=1−6x2 ⇒ 6x2+5x−1=0.
Factor: (6x−1)(x+1)=0⇒x=61 or x=−1. For x=−1 both angles are negative, so their sum cannot be 4π; reject it. Hence x=61.
Part (ii). f(x)=x2−4x−3 is a polynomial, so continuous on [1,4] and differentiable on (1,4) — the MVT hypotheses hold. …
Showing the 12 most recent of 39 on this concept.
- CBSE 2026Set A1 markMCQQ.tan−1(−31)=(a) 3π(b) 6π(c) −3π(d) −6π
›Reveal solutionSolution
tan−1(−31)=−6π.
The principal value of tan−1 lies in (−2π,2π).
We need the angle θ in this range with tanθ=−31.
…
- CBSE 2026Set A1 markMCQQ.2tan−131=(a) tan−123(b) tan−143(c) tan−134(d) tan−132
›Reveal solutionSolution
2tan−131=tan−143.
Use the double-angle identity valid for ∣x∣<1:
2tan−1x=tan−11−x22x.
Here x=31: …
- CBSE 2026Set A1 markMCQQ.x∈R, cot(tan−1x+cot−1x)=(a) 1(b) 21(c) 0(d) 31
›Reveal solutionSolution
cot(tan−1x+cot−1x)=cot2π=0.
For all real x, the complementary identity gives
tan−1x+cot−1x=2π.
Therefore …
- CBSE 2026Set A1 markMCQQ.tan−12+tan−13=(a) −4π(b) 4π(c) 43π(d) π
›Reveal solutionSolution
tan−12+tan−13=43π.
Use the addition formula. With a=2, b=3 we have ab=6>1, so
tan−1a+tan−1b=π+tan−11−aba+b.
Compute:
1−aba+b=1−65=−55=−1.
So …
- CBSE 2026Set A1 markMCQQ.tan−1yx−tan−1x+yx−y=(a) −43π(b) 2π(c) 4π(d) 3π
›Reveal solutionSolution
tan−1yx−tan−1x+yx−y=4π.
Use tan−1a−tan−1b=tan−11+aba−b with a=yx, b=x+yx−y.
Numerator:
a−b=yx−x+yx−y=y(x+y)x(x+y)−y(x−y)=y(x+y)x2+xy−xy+y2=y(x+y)x2+y2.
Denominator: …
- CBSE 2026Set A1 markMCQQ.∣x∣≤1, cos−1(1+x21−x2)=(a) 2cos−1x(b) 2sin−1x(c) 2tan−1x(d) tan−12x
›Reveal solutionSolution
cos−11+x21−x2=2tan−1x (for 0≤x≤1).
Put x=tanθ, so θ=tan−1x. Then
1+x21−x2=1+tan2θ1−tan2θ=cos2θ.
Therefore …
- CBSE 2025Set E1 markMCQQ.cot−1(tan7π)=(a) 7π(b) 145π(c) 149π(d) 143π
›Reveal solutionSolution
Convert the tangent to a cotangent using complementary angles; the answer is 145π.
Use tanθ=cot(2π−θ):
tan7π=cot(2π−7π)=cot147π−2π=cot145π.
…
- CBSE 2025Set E1 markMCQQ.tan−1(−3)=(a) 6π(b) 3π(c) 32π(d) −3π
›Reveal solutionSolution
tan−1 is an odd function with principal range (−2π,2π); the value is −3π.
Since tan3π=3 and tan−1(−x)=−tan−1x, …
- CBSE 2025Set E1 markMCQQ.tan−1(3)−cot−1(−3)=(a) 0(b) −2π(c) π(d) 2π
›Reveal solutionSolution
Evaluate each inverse function in its principal range and subtract; result −2π.
First, tan−1(3)=3π.
For cot−1(−3), the principal range of cot−1 is (0,π). We need cotθ=−3 with θ∈(0,π). Since cot6π=3,
cot−1(−3)=π−6π=65π.
…
- CBSE 2025Set E1 markMCQQ.tan−121+tan−131=(a) π(b) 4π(c) 2π(d) 3π
›Reveal solutionSolution
Use the sum formula for inverse tangents; the sum is 4π.
When xy<1, tan−1x+tan−1y=tan−11−xyx+y. Here xy=61<1, so …
- CBSE 2025Set E1 markMCQQ.tan{21(tan−1x+tan−1x1)}=(a) 1(b) 3(c) 0(d) ∞
›Reveal solutionSolution
tan−1x+tan−1x1=2π; half is 4π; tan4π=1.
For x>0 there is a standard identity:
tan−1x+tan−1x1=2π.
Taking half: …
- CBSE 2025Set ANNUAL1 markMCQQ.For what value of x, 2tan−1x=sin−1(23)?(i) 31(ii) 31(iii) 41(iv) 51
›Reveal solutionSolution
Simplify the RHS first, then solve for x.
sin−1(23)=3π
So the equation becomes: …
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