Advanced Proofs of Trigonometric Identities for CBSE Class 10

In school mathematics, few questions appear more daunting.

Important: <u>When proving trigonometric identities, always manipulate the more complex side (usually LHS) step-by-step to arrive at the other side.</u> at first glance than complex trigonometric identity proofs involving nested square roots, cubic powers, and multi-tier fractions. Yet, behind their intimidating appearance, every identity proof is governed by a small set of repeatable algebraic patterns.

In CBSE Class 10 Mathematics, Section C and Section D consistently feature 4-mark and 5-mark identity proof questions. Mastering the four core problem-solving strategies—sine-cosine conversion, algebraic factoring, conjugate rationalization, and strategic substitution of 1—gives students the structured confidence to solve any identity on the board exam.

What You Will Learn

• The mental framework for identity proofs: LHS vs. RHS strategy
• Strategy 1: Converting all terms to basic $\sin \theta$ and $\cos \theta$
• Strategy 2: Algebraic factoring identities ($a^2 - b^2, a^3 \pm b^3, (a \pm b)^2$)
• Strategy 3: Conjugate rationalization of denominators containing $(1 \pm \sin \theta)$ or $(1 \pm \cos \theta)$
• Strategy 4: Strategic substitution of the number $1$ by $(\sec^2 \theta - \tan^2 \theta)$ or $(\csc^2 \theta - \cot^2 \theta)$
• Complete, step-by-step proofs of five classic CBSE board examination identities
• Presentation rules to ensure maximum step marks

1. The Four Master Strategies for Proving Identities

When confronted with an identity proof:

                           The 4 Master Strategies
                                      |
       +------------------+-----------+-----------+------------------+
       |                  |                       |                  |
Strategy 1            Strategy 2              Strategy 3         Strategy 4
Convert to Sin & Cos  Algebraic Factoring     Conjugate          Substitute "1"
(tan → sin/cos)       (a² - b², a³ ± b³)      Rationalization    with sec² - tan²
                      (Take common terms)     (under radicals)   or csc² - cot²

1. Strategy 1 (Convert to Sine and Cosine): If an identity involves a mix of $\tan, \cot, \sec,$ and $\csc$, rewrite every single term in terms of $\sin$ and $\cos$: $\tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}$
2. Strategy 2 (Algebraic Factoring): Use standard algebraic identities to factor expressions and cancel terms: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ $a^2 - b^2 = (a - b)(a + b)$
3. Strategy 3 (Conjugate Rationalization): Whenever an expression features a radical like $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$ or a denominator like $(1 - \cos A)$, multiply both numerator and denominator by its conjugate $(1 + \cos A)$. This creates $(1 - \cos^2 A) = \sin^2 A$, eliminating the binomial denominator!
4. Strategy 4 (Strategic Substitution of 1): In difficult problems where you need to introduce secant and tangent, replace the number $1$ with $(\sec^2 \theta - \tan^2 \theta)$ or $(\csc^2 \theta - \cot^2 \theta)$.

2. Solved Classic CBSE Board Exam Identities

Identity 1: The Radical Conjugate Proof (NCERT Classic)

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

Proof:

1. Start with the Left-Hand Side (LHS): $\text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$
2. Multiply numerator and denominator inside the square root by the conjugate $(1 + \sin A)$: $\text{LHS} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}}$
3. Simplify numerator and denominator: $\text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$
4. Apply identity $1 - \sin^2 A = \cos^2 A$: $\text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A}$
5. Split the fraction: $\text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \mathbf{\sec A + \tan A} = \text{RHS}$
6. Hence, proved.

Identity 2: Factoring Cubic Terms (CBSE Board Question)

Problem: Prove that: $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$

Proof:

1. Start with LHS and factor out common terms:
   • In the numerator, take $\sin \theta$ common: $\sin \theta(1 - 2\sin^2 \theta)$
   • In the denominator, take $\cos \theta$ common: $\cos \theta(2\cos^2 \theta - 1)$ $\text{LHS} = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)} = \tan \theta \cdot \left[ \frac{1 - 2\sin^2 \theta}{2\cos^2 \theta - 1} \right]$
2. Use identity $\sin^2 \theta = 1 - \cos^2 \theta$ in the numerator: $1 - 2\sin^2 \theta = 1 - 2(1 - \cos^2 \theta) = 1 - 2 + 2\cos^2 \theta = 2\cos^2 \theta - 1$
3. Substitute back into the fraction: $\text{LHS} = \tan \theta \cdot \left[ \frac{2\cos^2 \theta - 1}{2\cos^2 \theta - 1} \right] = \tan \theta \cdot (1) = \mathbf{\tan \theta} = \text{RHS}$
4. Hence, proved.

Identity 3: The Expanding Squares Identity (NCERT Classic)

Problem: Prove that: $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

Proof:

1. Expand both squared binomials using $(a + b)^2 = a^2 + 2ab + b^2$: $\text{LHS} = (\sin^2 A + 2\sin A \csc A + \csc^2 A) + (\cos^2 A + 2\cos A \sec A + \sec^2 A)$
2. Use reciprocal relations $\sin A \csc A = 1$ and $\cos A \sec A = 1$: $\text{LHS} = \sin^2 A + 2(1) + \csc^2 A + \cos^2 A + 2(1) + \sec^2 A$ $\text{LHS} = (\sin^2 A + \cos^2 A) + 2 + 2 + \csc^2 A + \sec^2 A$
3. Substitute $\sin^2 A + \cos^2 A = 1$: $\text{LHS} = 1 + 4 + \csc^2 A + \sec^2 A = 5 + \csc^2 A + \sec^2 A$
4. Use identities $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$: $\text{LHS} = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)$ $\text{LHS} = (5 + 1 + 1) + \tan^2 A + \cot^2 A = \mathbf{7 + \tan^2 A + \cot^2 A} = \text{RHS}$
5. Hence, proved.

Identity 4: The Benchmark 5-Mark Identity (CBSE Classic)

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

Proof:

1. Divide numerator and denominator by $\sin A$ to introduce $\cot A$ and $\csc A$: $\text{LHS} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}$
2. Rearrange terms in the numerator: $\text{LHS} = \frac{(\cot A + \csc A) - 1}{(\cot A - \csc A + 1)}$
3. Apply Strategy 4: Replace $1$ in the numerator with $(\csc^2 A - \cot^2 A)$: $\text{LHS} = \frac{(\csc A + \cot A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1}$
4. Factor $(\csc^2 A - \cot^2 A)$ as $(\csc A + \cot A)(\csc A - \cot A)$: $\text{LHS} = \frac{(\csc A + \cot A) - [(\csc A + \cot A)(\csc A - \cot A)]}{\cot A - \csc A + 1}$
5. Factor out the 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. Notice that $[1 - \csc A + \cot A]$ and $[\cot A - \csc A + 1]$ are completely identical and cancel each other out! $\text{LHS} = \mathbf{\csc A + \cot A} = \text{RHS}$
7. Hence, proved.

3. Summary and Examination Tips

Exam Tip: Always write "LHS = ..." at the beginning and conclude with "= RHS, Hence Proved". If an identity gets messy, simplify LHS and RHS separately to the same intermediate expression!

Common Mistake: Cancelling terms across addition or subtraction. In $\frac{\sin A + \cos A}{\sin A}$, you CANNOT cross out $\sin A$ to get $1 + \cos A$! You can only cancel common factors that multiply the entire numerator and denominator.