On this page
Some trigonometric identities are true only under a condition, most often that A, B and C are the angles of a triangle. They look like a separate topic with formulas to memorise, but every one of them comes from a single substitution. This post shows that substitution, proves three standard results with numerical checks, and lists the places where the condition quietly fails.
Are conditional identities in the syllabus?
Not by name. NCERT's Class 11 chapter Trigonometric Functions (2026-27 reprint) has no section on them, and the JEE (Main) 2026 syllabus lists "Trigonometrical identities and trigonometrical functions" in Unit 14. The JEE (Advanced) 2026 brochure lists addition and subtraction formulae and formulae for multiple and sub-multiple angles. Conditional identities are applications of exactly these formulas, so you can prove every one of them with nothing outside either syllabus.
NCERT already contains the idea. Its Example 14 proves tan 3x tan 2x tan x = tan 3x − tan 2x − tan x by writing tan 3x = tan(2x + x) and cross-multiplying. Replace 3x by π − C, and you get the first identity below.
What is the one substitution?
If A + B + C = π, then A + B = π − C. Using the allied-angle results:
sin(A + B) = sin C
cos(A + B) = −cos C
tan(A + B) = −tan C
For half-angles, (A + B)/2 = π/2 − C/2, so sin((A + B)/2) = cos(C/2), cos((A + B)/2) = sin(C/2) and tan((A + B)/2) = cot(C/2).
Worked example 1: tan A + tan B + tan C = tan A tan B tan C
tan(A + B) = −tan C, so (tan A + tan B) ÷ (1 − tan A tan B) = −tan C.
Cross-multiply: tan A + tan B = −tan C + tan A tan B tan C.
Rearrange: tan A + tan B + tan C = tan A tan B tan C.
The condition hides an exception: none of the angles can be π/2, or a tangent is undefined.
Numerical check with A = 50°, B = 60°, C = 70°: both sides come to 5.671.
A use: in a triangle with tan A = 1 and tan B = 2, find tan C. From step 1, tan C = −(1 + 2) ÷ (1 − 2) = 3. Check: 1 + 2 + 3 = 6 and 1 × 2 × 3 = 6. The angles are 45°, about 63.43° and about 71.57°, which add to 180°.
Worked example 2: sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Pair the first two: sin 2A + sin 2B = 2 sin(A + B) cos(A − B) = 2 sin C cos(A − B).
Write the third with the substitution: sin 2C = 2 sin C cos C = −2 sin C cos(A + B).
Add: 2 sin C [cos(A − B) − cos(A + B)].
Use cos(A − B) − cos(A + B) = 2 sin A sin B: the total is 4 sin A sin B sin C.
Numerical check with 50°, 60°, 70°: both sides come to 2.494.
Worked example 3: the half-angle identity
Prove tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = 1.
tan((A + B)/2) = cot(C/2) = 1 ÷ tan(C/2).
Expand the left side: (tan(A/2) + tan(B/2)) ÷ (1 − tan(A/2) tan(B/2)) = 1 ÷ tan(C/2).
Cross-multiply: tan(C/2) tan(A/2) + tan(C/2) tan(B/2) = 1 − tan(A/2) tan(B/2).
Rearrange to get the result.
Numerical check with 50°, 60°, 70°: the sum is 1.000. The same argument shows that if A + B + C = π/2, then tan A tan B + tan B tan C + tan C tan A = 1.
Which slips cost marks?
Sign of cos(A + B). It is −cos C, not cos C.
Proving by substituting numbers. A numerical check catches errors but proves nothing; a proof must hold for every triangle.
Ignoring the condition. These identities are false for general angles. Try A = B = C = 30°.
Expanding everything. Writing sin 2A as 2 sin A cos A for all three terms first leads to a page of algebra. Pair, substitute, then expand.
When a proof stalls halfway, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, which helps you see which pairing the solution chose.
A practice routine
Prove cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C.
Prove cos²A + cos²B + cos²C + 2 cos A cos B cos C = 1.
Check both with A = 50°, B = 60°, C = 70° on a calculator.
Prove NCERT's Example 14 without looking, then redo it with π − C in place of 3x.
If allied angles such as sin(π − x) still need a lookup, revise them first with our post on allied angles in Class 11 trigonometry. Tonight, prove example 1 from a blank page and do the 50°, 60°, 70° check.
