Math Lab
Home/Class X/Ch 8/Trigonometric identities

Trigonometric identities

A trigonometric identity is an equation involving trig ratios that holds for all (valid) values of the angle. Three identities , all springing from the Pythagorean theorem , are the foundation.

The three Pythagorean identities

For any acute θ\theta: sin⁡2θ+cos⁡2θ=1,\sin^2\theta + \cos^2\theta = 1, 1+tan⁡2θ=sec⁡2θ,1 + \tan^2\theta = \sec^2\theta, 1+cot⁡2θ=csc⁡2θ.1 + \cot^2\theta = \csc^2\theta.

Derivation

In a right triangle with θ\theta, write opp =a= a, adj =b= b, hyp =c= c. By Pythagoras, a2+b2=c2a^2 + b^2 = c^2.

Divide by c2c^2: a2c2+b2c2=1  ⟹  sin⁡2θ+cos⁡2θ=1.\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1 \implies \sin^2\theta + \cos^2\theta = 1.

Divide by b2b^2: a2b2+1=c2b2  ⟹  tan⁡2θ+1=sec⁡2θ.\frac{a^2}{b^2} + 1 = \frac{c^2}{b^2} \implies \tan^2\theta + 1 = \sec^2\theta.

Divide by a2a^2: 1+b2a2=c2a2  ⟹  1+cot⁡2θ=csc⁡2θ.1 + \frac{b^2}{a^2} = \frac{c^2}{a^2} \implies 1 + \cot^2\theta = \csc^2\theta.

So all three are the same theorem , Pythagoras , viewed through different "lenses" of the right triangle.

Strategies for proving identities

A trigonometric identity proof looks like algebra:

  1. Pick one side (usually the messier one) and transform it step by step.
  2. Replace using known identities , sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 is the most useful.
  3. Convert everything to sin⁡\sin and cos⁡\cos if you get stuck. This is the "nuclear option" , it always works but can be slow.
  4. Factor and cancel , many identities boil down to common-factor extraction.
  5. Cross-multiply if both sides are fractions.

A few useful rewrites:

  • sec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta.
  • csc⁡2θ−1=cot⁡2θ\csc^2\theta - 1 = \cot^2\theta.
  • sin⁡θ⋅csc⁡θ=1\sin\theta \cdot \csc\theta = 1, cos⁡θ⋅sec⁡θ=1\cos\theta \cdot \sec\theta = 1, tan⁡θ⋅cot⁡θ=1\tan\theta \cdot \cot\theta = 1.
  • 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta, 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.

Worked examples

Example 1. Prove sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ\dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{1 - \cos\theta}{\sin\theta}.

Cross-multiply: sin⁡2θ=(1−cos⁡θ)(1+cos⁡θ)=1−cos⁡2θ\sin^2\theta = (1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta. ✓ (Pythagorean identity.) ■\blacksquare

Example 2. Prove tan⁡θ+cot⁡θ=sec⁡θ⋅csc⁡θ\tan\theta + \cot\theta = \sec\theta \cdot \csc\theta.

LHS =sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ=csc⁡θ⋅sec⁡θ= \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} = \dfrac{1}{\sin\theta \cos\theta} = \csc\theta \cdot \sec\theta. ■\blacksquare

Example 3. Prove (sin⁡θ+csc⁡θ)2+(cos⁡θ+sec⁡θ)2=7+tan⁡2θ+cot⁡2θ(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta.

Expand LHS: =sin⁡2+2sin⁡csc⁡+csc⁡2+cos⁡2+2cos⁡sec⁡+sec⁡2= \sin^2 + 2 \sin\csc + \csc^2 + \cos^2 + 2 \cos\sec + \sec^2 =(sin⁡2+cos⁡2)+2⋅1+2⋅1+csc⁡2+sec⁡2= (\sin^2 + \cos^2) + 2 \cdot 1 + 2 \cdot 1 + \csc^2 + \sec^2 =1+4+csc⁡2+sec⁡2= 1 + 4 + \csc^2 + \sec^2 =5+(1+cot⁡2)+(1+tan⁡2)= 5 + (1 + \cot^2) + (1 + \tan^2) =7+tan⁡2+cot⁡2= 7 + \tan^2 + \cot^2. ■\blacksquare

Example 4. Prove 1+tan⁡2A1+cot⁡2A=tan⁡2A\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A.

Numerator: 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A. Denominator: 1+cot⁡2A=csc⁡2A1 + \cot^2 A = \csc^2 A.

Ratio: sec⁡2A/csc⁡2A=(1/cos⁡2A)/(1/sin⁡2A)=sin⁡2A/cos⁡2A=tan⁡2A\sec^2 A / \csc^2 A = (1/\cos^2 A) / (1/\sin^2 A) = \sin^2 A / \cos^2 A = \tan^2 A. ■\blacksquare

Example 5. Prove cos⁡A1−sin⁡A+cos⁡A1+sin⁡A=2sec⁡A\dfrac{\cos A}{1 - \sin A} + \dfrac{\cos A}{1 + \sin A} = 2 \sec A.

LHS =cos⁡A[11−sin⁡A+11+sin⁡A]=cos⁡A⋅(1+sin⁡A)+(1−sin⁡A)1−sin⁡2A=cos⁡A⋅2cos⁡2A=2cos⁡A=2sec⁡A= \cos A \left[\dfrac{1}{1 - \sin A} + \dfrac{1}{1 + \sin A}\right] = \cos A \cdot \dfrac{(1 + \sin A) + (1 - \sin A)}{1 - \sin^2 A} = \cos A \cdot \dfrac{2}{\cos^2 A} = \dfrac{2}{\cos A} = 2 \sec A. ■\blacksquare

Try it yourself

  1. Prove sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 using a right triangle.
  2. Prove sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.
  3. Prove 1−sin⁡θcos⁡θ=cos⁡θ1+sin⁡θ\dfrac{1 - \sin\theta}{\cos\theta} = \dfrac{\cos\theta}{1 + \sin\theta}.
  4. Prove (1+tan⁡2θ)(1−sin⁡θ)(1+sin⁡θ)=1(1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) = 1.
  5. Prove tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+tan⁡θ+cot⁡θ\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta.
  6. Prove sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin\theta - 2\sin^3\theta}{2 \cos^3\theta - \cos\theta} = \tan\theta.
  7. Prove sec⁡4θ−sec⁡2θ=tan⁡2θ+tan⁡4θ\sec^4\theta - \sec^2\theta = \tan^2\theta + \tan^4\theta.
  8. Prove (sin⁡A+cos⁡A)(sec⁡A+csc⁡A)=2+sec⁡Acsc⁡A(\sin A + \cos A)(\sec A + \csc A) = 2 + \sec A \csc A.
  9. Prove cos⁡θ−sin⁡θ+1cos⁡θ+sin⁡θ−1=csc⁡θ+cot⁡θ\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} = \csc\theta + \cot\theta. (Hint: multiply numerator and denominator by cos⁡θ+sin⁡θ+1\cos\theta + \sin\theta + 1.)
  10. Prove (1+cot⁡θ−csc⁡θ)(1+tan⁡θ+sec⁡θ)=2(1 + \cot\theta - \csc\theta)(1 + \tan\theta + \sec\theta) = 2.

Pitfalls / Insight

  • sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2).
  • Don't divide by something that might be zero. Be wary of sin⁡θ−1,cos⁡θ\sin\theta - 1, \cos\theta in denominators.
  • One side at a time. Pick LHS or RHS, transform, end at the other side. Don't manipulate both at once.

Insight. Every classical identity is, at heart, Pythagoras divided by something. Recognise this and proofs become routine.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Identities
6 questions · pick the best answer
Q1

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

Q2

1+tan⁡2θ=1 + \tan^2\theta = :

Q3

1+cot⁡2θ=1 + \cot^2\theta = :

Q4

tan⁡θ+cot⁡θ=\tan\theta + \cot\theta = :

Q5

sec⁡2θ+csc⁡2θ=\sec^2\theta + \csc^2\theta = :

Q6

(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2: