How to prove trigonometric identities — strategy and common tricks

Question

Given a trigonometric identity to prove, what is the systematic strategy to approach it?

Solution — Step by Step

Pick the more complicated side (usually the LHS) and transform it step-by-step until it equals the simpler side. Never work on both sides simultaneously and “meet in the middle” — examiners deduct marks for this approach in CBSE boards.

The most reliable first move: replace all $\tan$ , $\cot$ , $\sec$ , $\cosec$ with their definitions in terms of $\sin$ and $\cos$ .

\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \cosec\theta = \frac{1}{\sin\theta}

After substitution, combine fractions, simplify, and use $\sin^2\theta + \cos^2\theta = 1$ .

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

1 + \tan^2\theta = \sec^2\theta

1 + \cot^2\theta = \cosec^2\theta

a^2 - b^2 = (a+b)(a-b) \text{ (algebraic, but used constantly)}

When you see $1 - \sin^2\theta$ , replace with $\cos^2\theta$ . When you see $\sec^2\theta - 1$ , replace with $\tan^2\theta$ . Pattern recognition speeds things up enormously.

When the expression has $(1 - \cos\theta)$ or $(1 + \sin\theta)$ in a denominator, multiply numerator and denominator by the conjugate:

\frac{1}{1 - \cos\theta} \times \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{1 + \cos\theta}{\sin^2\theta}

This is the single most useful trick for CBSE Class 10 identity proofs.

Prove: $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\cosec\theta$

Take LHS. Combine fractions (common denominator = $\sin\theta(1 + \cos\theta)$ ):

\text{LHS} = \frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}

Expand numerator: $\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1 + \cos\theta)$

\text{LHS} = \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \frac{2}{\sin\theta} = 2\cosec\theta = \text{RHS}

flowchart TD
    A["Trig Identity to Prove"] --> B["Pick the more complex side"]
    B --> C["Convert tan, sec, cot, cosec to sin/cos"]
    C --> D{"Single fraction or sum?"}
    D -->|"Sum of fractions"| E["Take LCM, combine"]
    D -->|"Single fraction"| F["Simplify numerator and denominator"]
    E --> G["Use sin2 + cos2 = 1 to simplify"]
    F --> G
    G --> H{"Stuck?"}
    H -->|"Yes"| I["Try conjugate multiplication"]
    H -->|"Yes"| J["Try factoring as a2 - b2"]
    H -->|"No"| K["Arrive at RHS"]

Why This Works

Every trigonometric function is ultimately a ratio of sides of a right triangle, and $\sin^2 + \cos^2 = 1$ encodes the Pythagorean theorem. All trig identities are algebraic consequences of this one fundamental relationship. Converting to sin and cos puts everything on the same footing, and then it is pure algebra.

Alternative Method

For some identities, substituting a specific angle (like $\theta = 45°$ ) first confirms the identity is true, which boosts confidence. Then use the LHS-to-RHS approach for the general proof. This “verify then prove” method reduces careless errors.

Common Mistake

The most penalized error in CBSE board exams: working on both LHS and RHS simultaneously and showing they are equal at some middle step. This is not a valid proof because you are assuming the result to prove the result (circular reasoning). Always start from one side and arrive at the other. If the examiner’s marking scheme says “deduct 1 mark for incorrect method,” this is the method they mean.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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