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NDA Maths · Properties of Triangle

Triangle Identities — A+B+C = π, Half & Double Angle

Because the three angles of a triangle add to π, the usual trig identities collapse into special triangle forms — sin(B+C) becomes sin A, and tan A + tan B + tan C becomes their product.

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Why this matters

14 PYQs, 6 HARD. Every identity here is a consequence of A + B + C = π. Knowing the handful of derived forms — the half-angle complements, the tan-product identity, and the cos 2A sum — turns intimidating expressions into one-line simplifications.

Concept 1 of 4

Consequences of A + B + C = π

Intuition

Since any one angle is π minus the other two, every sine/cosine of a sum of two angles rewrites in terms of the third — and the half-angle of two angles becomes the complementary half-angle of the third.

Definition

From A+B+C=πA + B + C = \pi:

  • sin⁡(B+C)=sin⁡A\sin(B+C) = \sin A, cos⁡(B+C)=−cos⁡A\cos(B+C) = -\cos A.
  • B+C2=π2−A2\dfrac{B+C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}, so sin⁡B+C2=cos⁡A2\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2} and tan⁡B+C2=cot⁡A2\tan\dfrac{B+C}{2} = \cot\dfrac{A}{2}.
  • Equations like sin⁡A=cos⁡B+cos⁡C\sin A = \cos B + \cos C are solved by replacing sin⁡A=sin⁡(B+C)\sin A = \sin(B+C) and using sum-to-product.

Half-angle complement

sin⁡B+C2=cos⁡A2,sin⁡(B+C)=sin⁡A\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2}, \qquad \sin(B+C) = \sin A

Worked example

In △ABC\triangle ABC, simplify cos⁡A+B2\cos\dfrac{A+B}{2}.
  1. A+B=π−CA + B = \pi - C, so A+B2=π2−C2\dfrac{A+B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2}.
  2. cos⁡ ⁣(π2−C2)=sin⁡C2\cos\!\left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \sin\dfrac{C}{2}.
Answer:cos⁡A+B2=sin⁡C2\cos\dfrac{A+B}{2} = \sin\dfrac{C}{2}.

From the bank · past-year question

Example 1Properties of TriangleMODERATE
Consider the following for triangle ABCABC: 1. sin⁡ ⁣(B+C2)=cos⁡ ⁣(A2)\sin\!\left(\dfrac{B+C}{2}\right) = \cos\!\left(\dfrac{A}{2}\right) 2. tan⁡ ⁣(B+C2)=cot⁡ ⁣(A2)\tan\!\left(\dfrac{B+C}{2}\right) = \cot\!\left(\dfrac{A}{2}\right) 3. sin⁡(B+C)=cos⁡A\sin(B+C) = \cos A 4. tan⁡(B+C)=−cot⁡A\tan(B+C) = -\cot A Which of the above are correct?

[Q49 · Apr · 2017]

Drill 3 more on consequences of a + b + c = π

Concept 2 of 4

The tan-Sum = tan-Product Identity

Intuition

In a triangle the sum of the three tangents equals their product — a surprising identity that turns a sum into a product (and vice versa). The cotangents satisfy a companion identity.

Definition

For A+B+C=πA + B + C = \pi:

  • tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C.
  • cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1\cot A\cot B + \cot B\cot C + \cot C\cot A = 1.
  • tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1\tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1.

Sign reading: cot⁡Acot⁡Bcot⁡C>0\cot A\cot B\cot C > 0 forces all angles acute (acute triangle).

tan sum = tan product

tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C

Worked example

In △ABC\triangle ABC, tan⁡A+tan⁡B+tan⁡C=6\tan A + \tan B + \tan C = 6. Find cot⁡Acot⁡Bcot⁡C\cot A\cot B\cot C.
  1. By the identity, tan⁡Atan⁡Btan⁡C=tan⁡A+tan⁡B+tan⁡C=6\tan A\tan B\tan C = \tan A + \tan B + \tan C = 6.
  2. cot⁡Acot⁡Bcot⁡C=1tan⁡Atan⁡Btan⁡C=16\cot A\cot B\cot C = \dfrac{1}{\tan A\tan B\tan C} = \dfrac{1}{6}.
Answer:16\dfrac{1}{6}.

From the bank · past-year question

Example 2Properties of TriangleMODERATE
In a triangle ABCABC, tan⁡A+tan⁡B+tan⁡C=k\tan A+\tan B+\tan C=k. What is the value of cot⁡Acot⁡Bcot⁡C\cot A\cot B\cot C?

[Q28 · Sep · 2024]

The sum equals the PRODUCT, only in a triangle

tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A\tan B\tan C holds because A+B+C=πA+B+C=\pi. It is NOT a general identity — don't apply it unless the three angles are a triangle's angles.
Drill 2 more on the tan-sum = tan-product identity

Concept 3 of 4

cos 2A Sums & Right-Angle Detection

Intuition

Sums of cos 2A or sin²A over the three angles collapse to clean values that signal the triangle's type — in particular, a specific value of these sums means one angle is exactly 90°.

Definition

Standard triangle identities:

  • cos⁡2A+cos⁡2B+cos⁡2C=−1−4cos⁡Acos⁡Bcos⁡C\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C.
  • sin⁡2A+sin⁡2B+sin⁡2C=2+2cos⁡Acos⁡Bcos⁡C\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A\cos B\cos C; it equals 2 iff the triangle is right-angled (one cosine is 0).
  • Equivalently cos⁡2A+cos⁡2B+cos⁡2C=1\cos^2 A + \cos^2 B + \cos^2 C = 1 iff right-angled. A value of cos⁡2A+cos⁡2B+cos⁡2C=−1\cos 2A + \cos 2B + \cos 2C = -1 forces cos⁡Acos⁡Bcos⁡C=0\cos A\cos B\cos C = 0.

Right-angle signature

sin⁡2A+sin⁡2B+sin⁡2C=2  ⟺  right-angled\sin^2 A + \sin^2 B + \sin^2 C = 2 \iff \text{right-angled}

Worked example

If △ABC\triangle ABC has sin⁡2A+sin⁡2B+sin⁡2C=2\sin^2 A + \sin^2 B + \sin^2 C = 2, what can you conclude?
  1. Use sin⁡2A+sin⁡2B+sin⁡2C=2+2cos⁡Acos⁡Bcos⁡C\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A\cos B\cos C.
  2. Setting it to 2 gives cos⁡Acos⁡Bcos⁡C=0\cos A\cos B\cos C = 0, so one cosine is 0 — that angle is 90∘90^\circ.
Answer:The triangle is right-angled.

From the bank · past-year question

Example 3Properties of TriangleHARD
Let ABCABC be a triangle. If cos⁡2A+cos⁡2B+cos⁡2C=−1\cos 2A+\cos 2B+\cos 2C=-1, then which one of the following is correct?

[Q41 · Sep · 2021]

Drill 1 more on cos 2a sums & right-angle detection

Concept 4 of 4

Half-Angle Formulas & Sum-to-Product

Intuition

The half-angle formulas express tan(A/2) through the sides and the inradius; sum-to-product turns cos A + cos B (and similar) into a product that simplifies using A + B + C = π. Both are workhorses for the harder identity questions.

Definition

  • Half-angle (sides): tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}, and sin⁡A2=(s−b)(s−c)bc\sin\dfrac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}.
  • Sum-to-product: cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}; in a triangle cos⁡A+B2=sin⁡C2\cos\dfrac{A+B}{2} = \sin\dfrac{C}{2}.
  • Product-to-sum: sin⁡Xsin⁡Y=12[cos⁡(X−Y)−cos⁡(X+Y)]\sin X\sin Y = \tfrac12[\cos(X-Y) - \cos(X+Y)], useful for sin⁡A2sin⁡3A2\sin\tfrac{A}{2}\sin\tfrac{3A}{2}-type expressions.
  • Note tan⁡B2+cot⁡B2=2sin⁡B\tan\dfrac{B}{2} + \cot\dfrac{B}{2} = \dfrac{2}{\sin B}.

Half-angle tangent

tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}

Worked example

In △ABC\triangle ABC with C=60∘C = 60^\circ, simplify cos⁡A+cos⁡Bcos⁡A−B2\dfrac{\cos A + \cos B}{\cos\frac{A-B}{2}}.
  1. Sum-to-product: cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}.
  2. Since C=60∘C = 60^\circ, A+B=120∘A + B = 120^\circ, so cos⁡A+B2=cos⁡60∘=12\cos\dfrac{A+B}{2} = \cos 60^\circ = \tfrac12.
  3. The expression is 2⋅12cos⁡A−B2÷cos⁡A−B22\cdot\tfrac12\cos\dfrac{A-B}{2} \div \cos\dfrac{A-B}{2}.
Answer:11.

From the bank · past-year question

Example 4Properties of TriangleMODERATE
If cos⁡A=34\cos A = \dfrac{3}{4}, then what is the value of sin⁡A2sin⁡3A2\sin\dfrac{A}{2}\sin\dfrac{3A}{2}?

[Q42 · Apr · 2019]

Drill 4 more on half-angle formulas & sum-to-product

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Consequences of A + B + C = π

    Half-angle complement

    sin⁡B+C2=cos⁡A2,sin⁡(B+C)=sin⁡A\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2}, \qquad \sin(B+C) = \sin A
  • The tan-Sum = tan-Product Identity

    tan sum = tan product

    tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C
  • cos 2A Sums & Right-Angle Detection

    Right-angle signature

    sin⁡2A+sin⁡2B+sin⁡2C=2  ⟺  right-angled\sin^2 A + \sin^2 B + \sin^2 C = 2 \iff \text{right-angled}
  • Half-Angle Formulas & Sum-to-Product

    Half-angle tangent

    tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}

Watch out for (1)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Properties of TriangleHARD
Consider the following statements: 1. If in a triangle ABC, A=2BA=2B and b=cb=c, then it must be an obtuse-angled triangle. 2. There exists no triangle ABC with A=40°A=40°, B=65°B=65° and ac=sin⁡40°csc⁡15°\frac{a}{c}=\sin 40°\csc 15°. Which of the above statements is/are correct?

[Q30 · Apr · 2020]

Example 2Properties of TriangleMODERATE
Consider the following statements: A. In a triangle ABCABC, if cot⁡A⋅cot⁡B⋅cot⁡C>0\cot A\cdot\cot B\cdot\cot C>0, then the triangle is an acute angled triangle. B. In a triangle ABCABC, if tan⁡A⋅tan⁡B⋅tan⁡C>0\tan A\cdot\tan B\cdot\tan C>0, then the triangle is an obtuse angled triangle. Which of the statements given above is/are correct?

[Q50 · Apr · 2024]

Example 3Properties of TriangleMODERATE
In a triangle ABCABC, AB=16AB=16 cm, BC=63BC=63 cm and AC=65AC=65 cm. What is the value of cos⁡2A+cos⁡2B+cos⁡2C\cos2A+\cos2B+\cos2C?

[Q46 · Apr · 2024]

Example 4Properties of TriangleHARD
In a triangle ABC, sin⁡A=cos⁡B+cos⁡C\sin A=\cos B+\cos C, then what is tan⁡ ⁣(B2)+cot⁡ ⁣(B2)\tan\!\left(\dfrac{B}{2}\right)+\cot\!\left(\dfrac{B}{2}\right) equal to?

[Q33 · Apr · 2026]

Example 5Properties of TriangleHARD
Consider the following statements: 1. If ABC is a right-angled triangle, right-angled at A and if sin⁡B=13\sin B=\frac{1}{3}, then csc⁡C=3\csc C=3. 2. If bcos⁡B=ccos⁡Cb\cos B=c\cos C and if the triangle ABC is not right-angled, then ABC must be isosceles. Which of the above statements is/are correct?

[Q29 · Apr · 2020]

Drill every past-year question on this subtopic

14 questions from the bank — paginated, with cart and Word-export support.

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