Q.If ∣x∣≤1, then 2tan−1x+sin−1(1+x22x) is equal to
(A) 4tan−1x
(B) 0
(C) 2π
(D) π
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Inverse Tangent Identity
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. …
Use the standard identity, valid exactly on the given domain.
For ∣x∣≤1,
sin−1(1+x22x)=2tan−1x.
(Reason: with x=tanθ, θ=tan−1x∈[−4π,4π], so 1+x22x=sin2θ and 2θ∈[−2π,2π] is already in the sin−1 rang …
On ∣x∣≤1 the identity sin−1(1+x22x)=2tan−1x holds, so the expression is 2tan−1x+2tan−1x=4tan−1x — option (A).
The idea
The fraction 1+x22x is exactly sin2θ when x=tanθ. The only subtlety is whether sin−1(sin2θ)=2θ, which needs 2θ inside the sin−1 range [−2π,2π].
Step 1 — Substitute
Let x=tanθ with θ=tan−1x. Since ∣x∣≤1, θ∈[−4π,4π]. Then
1+x22x=1+tan2θ2tanθ=sin2θ.
Step 2 — Peel off the sin−1
Because θ∈[−4π,4π], we have 2θ∈[−2π,2π], the principal range of sin−1. Hence …
Method: Applying standard 2tan−1x conversion identities with their domain conditions
Expressions such as sin−11+x22x can be rewritten as 2tan−1x — but only on the correct domain. The method is to identify the identity and verify its condition before using it.
Steps
Step 1: Recognise the standard form.
Memorise the trio
2tan−1x=sin−11+x22x=cos−11+x21−x2=tan−11−x22x,
each valid on its own interval.
Step 2: Check the domain condition for the specific piece.
sin−11+x22x=2tan−1x holds for ∣x∣≤1. (For x>1 it becomes π−2tan−1x; for x<−1, −π−2tan−1x.) Confirm the given domain matches before substituting. …
Common Mistakes
Mistake 1: Using sin−11+x22x=2tan−1x without checking ∣x∣≤1.
Why it's wrong: for x>1 the correct value is π−2tan−1x (and −π−2tan−1x for x<−1), so the clean identity fails outside [−1,1]. Correct approach: the given condition ∣x∣≤1 is exactly what makes sin−11+x22x=2tan−1x valid. …
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: …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.