Mastering Trigonometric Identities and Proofs: Class 10 Maths

In CBSE Class 10 Mathematics, no topic induces more test anxiety than proving trigonometric identities in Section C and Section D (carrying 3 to 4 marks). Unlike arithmetic or quadratic equations—where a mechanical algorithm always yields an answer—trigonometric proofs require creative algebraic foresight. Looking at an expression like $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$, many students freeze, unsure whether to square, factor, or divide.

However, trigonometric proofs are not random puzzles. Over $95\%$ of all board examination proofs can be dismantled using Five Systematic Strategic Rules.

In this master guide, we break down these 5 proven strategies and work through the most famous, high-weightage trigonometric proofs in the CBSE curriculum.

What You Will Learn

• The 3 fundamental Pythagorean trigonometric identities and their rearrangements
• Strategy 1: The Universal Sine-Cosine Conversion Technique
• Strategy 2: Conjugate Rationalization of Radical Expressions
• Strategy 3: Factoring via Algebraic Identities ($a^3 \pm b^3, a^4 - b^4$)
• Strategy 4: Independent Simplification of LHS and RHS
• Strategy 5: The Famous $\csc^2 A = 1 + \cot^2 A$ Division Technique (NCERT Exemplar)
• Step-by-step solutions to guaranteed board examination proofs

1. The Core Toolkit: Fundamental Identities

Before attempting any proof, you must have the three core Pythagorean identities and their algebraic transpositions memorized:

    Identity 1:   sin² θ + cos² θ = 1    ===>  sin² θ = 1 - cos² θ   OR   cos² θ = 1 - sin² θ
    Identity 2:   1 + tan² θ = sec² θ    ===>  sec² θ - tan² θ = 1   OR   tan² θ = sec² θ - 1
    Identity 3:   1 + cot² θ = csc² θ    ===>  csc² θ - cot² θ = 1   OR   cot² θ = csc² θ - 1

Reciprocal and Quotient Relations:

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

2. Strategy 1: The Universal Sine-Cosine Conversion

Rule of Thumb: When an identity contains a mixture of $\tan, \cot, \sec,$ and $\csc$, convert every single term into its fundamental $\sin$ and $\cos$ components!

Solved Example:

Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$

Proof:

1. Express LHS entirely in terms of $\sin \theta$ and $\cos \theta$: $\text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$
2. Simplify the denominators: $\text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$
3. Invert and multiply: $\text{LHS} = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}$
4. Make the denominators identical by factoring out a negative sign: $\text{LHS} = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}$
5. Take common denominator $\sin \theta \cos \theta (\sin \theta - \cos \theta)$: $\text{LHS} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
6. Apply the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $\text{LHS} = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
7. Cancel $(\sin \theta - \cos \theta)$ and substitute $\sin^2 \theta + \cos^2 \theta = 1$: $\text{LHS} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} = \mathbf{\sec \theta \csc \theta + 1} = \text{RHS}$ Hence Proved.

3. Strategy 2: Conjugate Rationalization for Radical Proofs

Rule of Thumb: Whenever a proof involves a square root over a fraction like $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$, multiply the numerator and denominator under the radical by the conjugate of the denominator!

Solved Example:

Prove that: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$

Proof:

1. Multiply numerator and denominator by the conjugate $(1 + \sin A)$: $\text{LHS} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$
2. Substitute $1 - \sin^2 A = \cos^2 A$: $\text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A}$
3. Split into two fractions: $\text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \mathbf{\sec A + \tan A} = \text{RHS}$ Hence Proved.

4. Strategy 3: The Famous Cot-Csc Identity Technique (NCERT Exemplar)

This specific problem is considered the hardest trigonometric proof in Class 10, appearing in Section D:

Prove that: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$, using the identity $\csc^2 A = 1 + \cot^2 A$.

Step-by-Step Solution:

1. Divide Every Term by $\sin A$: $\text{LHS} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}$
2. Group terms in the numerator: $\text{LHS} = \frac{(\cot A + \csc A) - 1}{\cot A - \csc A + 1}$
3. <u>Crucial Step: Replace the '1' in the numerator with $(\csc^2 A - \cot^2 A)$!</u> $\text{LHS} = \frac{(\csc A + \cot A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1}$
4. Factorize $(\csc^2 A - \cot^2 A) = (\csc A + \cot A)(\csc A - \cot A)$: $\text{LHS} = \frac{(\csc A + \cot A) - (\csc A + \cot A)(\csc A - \cot A)}{\cot A - \csc A + 1}$
5. Factor out common term $(\csc A + \cot A)$ from the numerator: $\text{LHS} = \frac{(\csc A + \cot A)[1 - (\csc A - \cot A)]}{\cot A - \csc A + 1}$ $\text{LHS} = \frac{(\csc A + \cot A)[1 - \csc A + \cot A]}{[\cot A - \csc A + 1]}$
6. The bracketed terms in numerator and denominator are identical and cancel completely: $\text{LHS} = \mathbf{\csc A + \cot A} = \text{RHS}$ Hence Proved.

5. Summary and Examination Tips

Exam Tip: If you get stuck halfway through transforming the LHS, STOP! Start simplifying the RHS on the next line. Often, both sides will reduce to the exact same intermediate expression (e.g., $\frac{1}{\sin A \cos A}$), proving $\text{LHS} = \text{RHS}$!

Common Mistake: Cancelling terms across addition signs, like cancelling $\sin A$ from $\frac{\sin A + \cos A}{\sin A}$. You can only cancel common factors that multiply the ENTIRE numerator and denominator!