The Fundamental Identity

An identity is an equation true for every value of the variable. The master trigonometric identity comes straight from the Pythagoras theorem:

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

Why: in a right triangle with opposite pp, adjacent bb, hypotenuse hh, Pythagoras gives p2+b2=h2p^2 + b^2 = h^2. Divide by h2h^2: (ph)2+(bh)2=1\left(\tfrac{p}{h}\right)^2 + \left(\tfrac{b}{h}\right)^2 = 1, i.e. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

Key Point: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 holds for all θ\theta. From it you can get sin⁡θ=1−cos⁡2θ\sin\theta = \sqrt{1 - \cos^2\theta} and vice versa.

[Board Important] Note sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2).

The Two Companion Identities

Dividing the master identity by cos⁡2θ\cos^2\theta and by sin⁡2θ\sin^2\theta gives two more:

1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

Derivation of the first: divide sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 by cos⁡2θ\cos^2\theta: tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta.

Key Point: The three identities are: sin⁡2+cos⁡2=1\sin^2+\cos^2=1, sec⁡2−tan⁡2=1\sec^2-\tan^2=1, csc⁡2−cot⁡2=1\csc^2-\cot^2=1.

[JEE/NEET Tip] Knowing all three forms (and their rearrangements like sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1) saves time in proofs.

Strategy for Proving Identities

To prove an identity (LHS = RHS):

  1. Start with the more complicated side and simplify it towards the other.
  2. Convert everything to sin⁡θ\sin\theta and cos⁡θ\cos\theta when stuck — they're the 'common currency'.
  3. Use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to replace or create terms.
  4. Look for conjugate multiplication: multiply 1sec⁡θ−tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} by sec⁡θ+tan⁡θsec⁡θ+tan⁡θ\dfrac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}, since sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.

Key Point: When stuck, convert to sines and cosines and bring to a common denominator.

[Board Important] Never cross-multiply across an unproven identity — work each side independently or transform one side into the other.

Useful Consequences

Some handy rearrangements you'll use constantly:

  • sin⁡2θ=1−cos⁡2θ=(1−cos⁡θ)(1+cos⁡θ)\sin^2\theta = 1 - \cos^2\theta = (1-\cos\theta)(1+\cos\theta).
  • sec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta and csc⁡2θ−1=cot⁡2θ\csc^2\theta - 1 = \cot^2\theta.
  • 1sec⁡θ−tan⁡θ=sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta (conjugates, product 1).

Key Point: The difference-of-squares factorisation 1−cos⁡2θ=(1−cos⁡θ)(1+cos⁡θ)1 - \cos^2\theta = (1-\cos\theta)(1+\cos\theta) unlocks many proofs.

[Board Important] Recognising a conjugate pair like (sec⁡θ−tan⁡θ)(\sec\theta - \tan\theta) and (sec⁡θ+tan⁡θ)(\sec\theta + \tan\theta) is a high-frequency exam skill.

Solved Examples

Example 1: Use the identity

If cos⁡θ=35\cos\theta = \tfrac35, find sin⁡θ\sin\theta using an identity.

Solution:

  1. sin⁡2θ=1−cos⁡2θ=1−925=1625\sin^2\theta = 1 - \cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}.
  2. sin⁡θ=45\sin\theta = \tfrac45 (positive, acute).

Final Answer: sin⁡θ=45\sin\theta = \tfrac45.

Takeaway: sin⁡θ=1−cos⁡2θ\sin\theta = \sqrt{1 - \cos^2\theta} for acute angles.

Example 2: Simplify

Simplify (1−sin⁡2θ)sec⁡2θ(1 - \sin^2\theta)\sec^2\theta.

Solution:

  1. 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta.
  2. cos⁡2θ⋅sec⁡2θ=cos⁡2θ⋅1cos⁡2θ=1\cos^2\theta \cdot \sec^2\theta = \cos^2\theta \cdot \dfrac{1}{\cos^2\theta} = 1.

Final Answer: 1.

Takeaway: Replace 1−sin⁡2θ1 - \sin^2\theta with cos⁡2θ\cos^2\theta.

Example 3: Prove an identity

Prove that sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ=2csc⁡θ\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = 2\csc\theta.

Solution:

  1. LHS =sin⁡2θ+(1+cos⁡θ)2sin⁡θ(1+cos⁡θ)= \dfrac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}.
  2. Numerator =sin⁡2θ+1+2cos⁡θ+cos⁡2θ=1+1+2cos⁡θ=2(1+cos⁡θ)= \sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = 1 + 1 + 2\cos\theta = 2(1+\cos\theta).
  3. LHS =2(1+cos⁡θ)sin⁡θ(1+cos⁡θ)=2sin⁡θ=2csc⁡θ= \dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta} = 2\csc\theta.

Final Answer: Proved.

Takeaway: Common denominator + sin⁡2+cos⁡2=1\sin^2+\cos^2=1 collapses the numerator.

Example 4: Conjugate trick

Prove 1sec⁡θ−tan⁡θ=sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta.

Solution:

  1. Multiply numerator and denominator by (sec⁡θ+tan⁡θ)(\sec\theta + \tan\theta).
  2. Denominator =sec⁡2θ−tan⁡2θ=1= \sec^2\theta - \tan^2\theta = 1.
  3. So LHS =sec⁡θ+tan⁡θ= \sec\theta + \tan\theta.

Final Answer: Proved.

Takeaway: sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 makes conjugates collapse.

Example 5: Evaluate using identity

If sec⁡θ=135\sec\theta = \tfrac{13}{5}, find tan⁡θ\tan\theta.

Solution:

  1. tan⁡2θ=sec⁡2θ−1=16925−1=14425\tan^2\theta = \sec^2\theta - 1 = \tfrac{169}{25} - 1 = \tfrac{144}{25}.
  2. tan⁡θ=125\tan\theta = \tfrac{12}{5}.

Final Answer: tan⁡θ=125\tan\theta = \tfrac{12}{5}.

Takeaway: tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1.

Example 6: Prove

Prove sin⁡4θ−cos⁡4θ=sin⁡2θ−cos⁡2θ\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta.

Solution:

  1. LHS =(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)= (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta).
  2. =(sin⁡2θ−cos⁡2θ)(1)=sin⁡2θ−cos⁡2θ= (\sin^2\theta - \cos^2\theta)(1) = \sin^2\theta - \cos^2\theta.

Final Answer: Proved.

Takeaway: Difference of squares + sin⁡2+cos⁡2=1\sin^2+\cos^2=1.

Example 7: Convert to sin/cos

Prove tan⁡θ+cot⁡θ=sec⁡θcsc⁡θ\tan\theta + \cot\theta = \sec\theta\csc\theta.

Solution:

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

Final Answer: Proved.

Takeaway: Convert to sin/cos, combine, then use the master identity.

Example 8: Eliminate the parameter

If x=asec⁡θx = a\sec\theta and y=btan⁡θy = b\tan\theta, show x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1.

Solution:

  1. x2a2=sec⁡2θ\dfrac{x^2}{a^2} = \sec^2\theta, y2b2=tan⁡2θ\dfrac{y^2}{b^2} = \tan^2\theta.
  2. sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.

Final Answer: Proved.

Takeaway: Eliminating θ\theta with sec⁡2−tan⁡2=1\sec^2 - \tan^2 = 1 gives a hyperbola.

Example 9: Harder proof

Prove 1+sin⁡θ1−sin⁡θ=(sec⁡θ+tan⁡θ)2\dfrac{1 + \sin\theta}{1 - \sin\theta} = (\sec\theta + \tan\theta)^2.

Solution:

  1. RHS =sec⁡2θ+2sec⁡θtan⁡θ+tan⁡2θ= \sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta.
  2. =1cos⁡2θ+2sin⁡θcos⁡2θ+sin⁡2θcos⁡2θ=1+2sin⁡θ+sin⁡2θcos⁡2θ= \dfrac{1}{\cos^2\theta} + \dfrac{2\sin\theta}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{1 + 2\sin\theta + \sin^2\theta}{\cos^2\theta}.
  3. =(1+sin⁡θ)21−sin⁡2θ=(1+sin⁡θ)2(1−sin⁡θ)(1+sin⁡θ)=1+sin⁡θ1−sin⁡θ= \dfrac{(1+\sin\theta)^2}{1 - \sin^2\theta} = \dfrac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)} = \dfrac{1+\sin\theta}{1-\sin\theta}.

Final Answer: Proved.

Takeaway: Expand the square, use cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta, then factor.

Example 10: Value of an expression

If sin⁡θ+cos⁡θ=2\sin\theta + \cos\theta = \sqrt2, find sin⁡θcos⁡θ\sin\theta\cos\theta.

Solution:

  1. Square both sides: sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=2\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 2.
  2. 1+2sin⁡θcos⁡θ=2⇒sin⁡θcos⁡θ=121 + 2\sin\theta\cos\theta = 2 \Rightarrow \sin\theta\cos\theta = \tfrac12.

Final Answer: sin⁡θcos⁡θ=12\sin\theta\cos\theta = \tfrac12.

Takeaway: Squaring a sum introduces the cross-term you want; use sin⁡2+cos⁡2=1\sin^2+\cos^2=1.