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Trigonometric identities: the five moves that solve almost all of them

7 min read

Trigonometric identity questions look like they need inspiration. You stare at two expressions that plainly do not resemble each other and are asked to show they are the same, and it feels like either you see it or you do not.

You do not need inspiration. Almost every identity in the Class 10 syllabus falls to the same five moves, tried in roughly the same order. Learn the order and these stop being a gamble.

Two rules before you start

Work on one side only. You are proving the two sides are equal, so you cannot assume it yet. That means no cross-multiplying, no adding the same thing to both sides, no squaring both sides. Pick a side, transform it, arrive at the other.

Start from the messier side. It is far easier to simplify a complicated expression than to complicate a simple one. If one side is a single term and the other is a fraction with three terms, you start at the fraction.

The five moves

Try them in this order. One of them almost always opens the question.

Move 1: convert everything to sin and cos

This is the default opening and it works more often than anything else. Every other trigonometric ratio is built from these two:

tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \cot\theta=\frac{\cos\theta}{\sin\theta}
sec⁡θ=1cos⁡θ,cosec⁡θ=1sin⁡θ\sec\theta=\frac{1}{\cos\theta},\quad \operatorname{cosec}\theta=\frac{1}{\sin\theta}

Once everything is in sin and cos, the expression is usually just algebra, and algebra you can do.

Move 2: reach for the Pythagorean identity

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

The trick is to remember it works in every direction. All of these are the same fact, and the useful one is whichever matches what is in front of you:

  • sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta
  • cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta
  • 1−sin⁡2θ=cos⁡2θ1-\sin^2\theta=\cos^2\theta

Whenever you see 1−cos⁡2θ1-\cos^2\theta or 1−sin⁡2θ1-\sin^2\theta anywhere, replace it immediately. That substitution is the single most common step in the whole chapter.

Move 3: combine fractions over a common denominator

If the messy side has two fractions added or subtracted, put them over one denominator. The numerator that results very often simplifies using move 2, which is exactly why the question was built that way.

Move 4: factorise, especially difference of squares

Class 8 algebra, doing heavy lifting here:

a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)

So 1−sin⁡2θ1-\sin^2\theta can be read as (1+sin⁡θ)(1−sin⁡θ)(1+\sin\theta)(1-\sin\theta). When the other side of the identity contains 1+sin⁡θ1+\sin\theta, that factorisation is the whole answer.

Move 5: multiply by the conjugate

The specialist move, for when you are stuck with a 1+cos⁡θ1+\cos\theta in a denominator and the other side wants 1−cos⁡θ1-\cos\theta. Multiply top and bottom by the conjugate:

11+cos⁡θ×1−cos⁡θ1−cos⁡θ=1−cos⁡θ1−cos⁡2θ=1−cos⁡θsin⁡2θ\begin{aligned} \frac{1}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta} &=\frac{1-\cos\theta}{1-\cos^2\theta} \\[6pt] &=\frac{1-\cos\theta}{\sin^2\theta} \end{aligned}

Notice the chain: conjugate creates a difference of squares, which becomes a Pythagorean substitution. Moves 5, 4 and 2 in three lines. When you see a conjugate on the other side of the identity, this is the intended route.

Putting it together: a complete worked proof

Let us prove the standard board question: sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ=2cosec⁡θ\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = 2\operatorname{cosec}\theta.

LHS=sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ\text{LHS} = \frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta}
Combine fractions over LCM sin⁡θ(1+cos⁡θ):LHS=sin⁡2θ+(1+cos⁡θ)2sin⁡θ(1+cos⁡θ)\text{Combine fractions over LCM } \sin\theta(1+\cos\theta): \text{LHS} = \frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}
Expand numerator using (a+b)2=a2+2ab+b2:LHS=sin⁡2θ+1+2cos⁡θ+cos⁡2θsin⁡θ(1+cos⁡θ)\text{Expand numerator using } (a+b)^2 = a^2+2ab+b^2: \text{LHS} = \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1+\cos\theta)}
Rearrange terms: LHS=(sin⁡2θ+cos⁡2θ)+1+2cos⁡θsin⁡θ(1+cos⁡θ)\text{Rearrange terms: } \text{LHS} = \frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1+\cos\theta)}
Apply Pythagorean identity sin⁡2θ+cos⁡2θ=1:LHS=1+1+2cos⁡θsin⁡θ(1+cos⁡θ)\text{Apply Pythagorean identity } \sin^2\theta+\cos^2\theta=1: \text{LHS} = \frac{1 + 1 + 2\cos\theta}{\sin\theta(1+\cos\theta)}
Add constants: LHS=2+2cos⁡θsin⁡θ(1+cos⁡θ)\text{Add constants: } \text{LHS} = \frac{2 + 2\cos\theta}{\sin\theta(1+\cos\theta)}
Factor out 2 in numerator: LHS=2(1+cos⁡θ)sin⁡θ(1+cos⁡θ)\text{Factor out } 2 \text{ in numerator: } \text{LHS} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}
Cancel common factor (1+cos⁡θ):LHS=2sin⁡θ\text{Cancel common factor } (1+\cos\theta): \text{LHS} = \frac{2}{\sin\theta}
Apply reciprocal identity 1sin⁡θ=cosec⁡θ:LHS=2cosec⁡θ\text{Apply reciprocal identity } \frac{1}{\sin\theta} = \operatorname{cosec}\theta: \text{LHS} = 2\operatorname{cosec}\theta
LHS=RHS(Hence Proved ✓)\text{LHS} = \text{RHS} \quad (\text{Hence Proved } \checkmark)

When you meet any identity, run this standard sequence:

  1. Look at both sides. Start on the messier one.
  2. Convert everything to sin and cos.
  3. If there are separate fractions, combine them.
  4. Hunt for sin⁡2\sin^2 or cos⁡2\cos^2 and apply the Pythagorean identity.
  5. Factorise anything factorisable.
  6. If you are stuck with a stubborn denominator, try the conjugate.
  7. Keep glancing at the target. It tells you which form to aim for.

That last point matters more than it sounds. The right-hand side is not just the destination, it is a hint. If the answer contains sec⁡θ\sec\theta, you know you are heading towards something over cos⁡θ\cos\theta. Read the target before you start.

Explore the ideas with YMB concept learning and guided practice. The Board Prep Pass includes all 14 Class 10 Maths chapters until 31 March 2027 for ₹999 one-time.

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Your turn

Next identity you meet, write down which of the five moves you tried and in what order, before solving. Which one opened it?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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