Q.If tanA=21, tanB=31, then tan(2A+B) is equal to
(A) 1
(B) 2
(C) 3
(D) 4
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Trigonometric Identity Proof
Trigonometric Identity Proof: From Intuition to Precision
Imagine you're standing at the corner of a right triangle. The two shorter sides — one horizontal, one vertical — and the sloping hypotenuse are all connected. If you change the angle at your corner, the lengths of the sides change, but the relationship between them stays fixed. That fixed relationship is what a trigonometric identity captures.
The Core Idea
A trigonometric identity is an equation involving trigonometric functions (like sinθ, cosθ, tanθ) that is true for every angle θ where both sides are defined. It's not a conditional equation (like sinθ=0.5, which is true only for specific angles). It's an eternal truth about how these functions relate.
The most famous one is:
sin2θ+cos2θ=1
This holds for any angle θ — acute, obtuse, negative, whatever. Why? Because on the unit circle, sinθ is the y-coordinate and cosθ is the x-coordinate of a point on a circle of radius 1. The Pythagorean theorem says x2+y2=1, so sin2θ+cos2θ=1 is just the Pythagorean theorem in disguise.
Proving an Identity: The Method
When you're asked to prove a trigonometric identity, you're not solving for an angle. You're showing that the left-hand side (LHS) and right-hand side (RHS) are the same expression, just written differently.
The golden rule: Start with one side and transform it into the other, using known identities and algebraic manipulation. Never move terms across the equals sign as if solving an equation — that assumes the identity is already true, which is what you're trying to prove.
A Simple Example
Prove: tanθ⋅cosθ=sinθ
Step 1: Pick a side to start with. Usually, the more complicated side is easier to simplify. Here, the LHS looks more complex.
Step 2: Replace tanθ with cosθsinθ (a known identity).
tanθ⋅cosθ=cosθsinθ⋅cosθ
Step 3: Cancel cosθ (provided cosθ=0 — but the identity holds for all angles where both sides are defined, and at cosθ=0, tanθ is undefined anyway).
=sinθ
That's it. The LHS simplifies exactly to the RHS.
The Toolbox of Known Identities
To prove any identity, you need to know the basic building blocks:
| Identity | Formula |
|---|---|
| Pythagorean | sin2θ+cos2θ=1 |
| Quotient | tanθ=cosθsinθ, cotθ=sinθcosθ |
| Reciprocal | cscθ=sinθ1, secθ=cosθ1, cotθ=tanθ1 |
| Even-Odd | sin(−θ)=−sinθ, cos(−θ)=cosθ |
A common mistake is to treat sin2θ as (sinθ)2 — which it is — but then incorrectly think sin2θ+cos2θ=1 means sinθ+cosθ=1. It does not. The square applies to the whole sine value, not to the angle.
A Slightly Harder Proof
Prove: cosθ1−cos2θ=sinθtanθ
Start with LHS: cosθ1−cos2θ
From the Pythagorean identity, 1−cos2θ=sin2θ. So:
cosθsin2θ=sinθ⋅cosθsinθ=sinθtanθ
That's the RHS. Done. …
Concept: Compound and double-angle formulas for tangent, with quadrant-independent evaluation.
Step 1 – Find tan2A
Using tan2A=1−tan2A2tanA:
tan2A=1−(21)22⋅21=1−411=431=34.
Step 2 – Apply tan(2A+B) formula
tan(2A+B)=1−tan2A⋅tanBtan2A+tanB=1−34⋅3134+31.
Step 3 – Simplify …
Compute tan2A=34 from the double-angle formula, then apply the sum formula with tanB=31 to get tan(2A+B)=3 — option (C).
Step 1 — tan2A. With tanA=21:
tan2A=1−tan2A2tanA=1−412⋅21=431=34
Step 2 — tan(2A+B). With tan2A=34 and tanB=31: …
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQQ.The value of sin15∘ is(a) 22+13(b) 22−13(c) 223−1(d) 223+1
›Reveal solutionSolution
Using sin(45°−30°), sin15° simplifies to (√3−1)/(2√2).
Write 15° = 45° − 30° and use sin(A−B) = sinA cosB − cosA sinB:
sin15∘=sin45∘cos30∘−cos45∘sin30∘=22⋅23−22⋅21=46−2
…
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markQ.Evaluate: sin50∘cos10∘+cos50∘sin10∘.
›Reveal solutionSolution
The expression matches the sin(A+B) identity exactly, giving sin60° = √3/2.
Recall the identity: sin(A+B)=sinAcosB+cosAsinB.
…
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 markQ.If tanx=43, find the value of cos2x.
›Reveal solutionSolution
With tanx=43, the identity cos2x=1+tan2x1−tan2x gives cos2x=257.
The double-angle identity for cosine in terms of tangent is
cos2x=1+tan2x1−tan2x
Substitute tanx=43, so tan2x=169:
…
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 markQ.Find the value of sin15°.
›Reveal solutionSolution
Using 15°=45°−30° and the identity sin(A−B)=sinAcosB−cosAsinB gives sin15°=46−2.
Write 15°=45°−30°. Then
sin15°=sin(45°−30°)=sin45°cos30°−cos45°sin30°.
Substituting the known values sin45°=cos45°=22, cos30°=23, sin30°=21: …
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 markQ.If cosx=54, find the value of cos2x.
›Reveal solutionSolution
Applying cos2x=2cos2x−1 with cosx=54 gives cos2x=257.
The double-angle formula for cosine (in terms of cosx alone) is
cos2x=2cos2x−1.
Substituting cosx=54: …
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL1 markQ.Find the value of 2sin15∘cos15∘.
›Reveal solutionSolution
Using the identity 2sinθcosθ=sin2θ, the expression collapses to sin30∘=21.
Recall the double-angle formula:
2sinθcosθ=sin2θ
Here θ=15∘, so …
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