Fundamental Trigonometric Identities for CBSE Class 10 Mathematics

In algebra, an identity is an equation that remains true for every possible numerical value substituted for its variables (such as $(a + b)^2 = a^2 + 2ab + b^2$). When an algebraic equation involves trigonometric ratios of an angle and holds true for all permissible angles, it is called a Trigonometric Identity.

In CBSE Class 10 Mathematics, Chapter 8 (Introduction to Trigonometry) derives the three fundamental Pythagorean trigonometric identities. These three equations form the master toolkit used to simplify complex trigonometric expressions, solve geometric equations, and prove high-weightage board exam identities.

What You Will Learn

• Definition and significance of a trigonometric identity
• Geometric derivation of the First Identity: $\sin^2 \theta + \cos^2 \theta = 1$
• Geometric derivation of the Second Identity: $1 + \tan^2 \theta = \sec^2 \theta$
• Geometric derivation of the Third Identity: $1 + \cot^2 \theta = \csc^2 \theta$
• Algebraic variations and the difference-of-squares reciprocal shortcut ($(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$)
• Expressing all trigonometric ratios in terms of a single ratio
• Solved CBSE board examination problems and common traps

1. Derivation of the Three Pythagorean Identities

All three fundamental identities originate from a single source.

Important: <u>A trigonometric identity is an equation that is true for all values of the acute angle theta for which the functions are defined.</u>: the Pythagoras Theorem applied to a right-angled triangle.

Consider a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$ and reference acute angle $\angle A = \theta$: $AB^2 + BC^2 = AC^2 \quad \text{--- (Pythagoras Theorem)}$

                       A (θ)
                       |                       |                     AB |  \ AC (Hypotenuse)
                       |                          +----+
                       B    C
                         BC

Derivation 1: The First Identity ($\sin^2 heta + \cos^2 heta = 1$)

Divide each term of the Pythagoras equation by $AC^2$ (the square of the hypotenuse): $\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}$ $\left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1$

Since $\frac{AB}{AC} = \cos A$ and $\frac{BC}{AC} = \sin A$, we obtain: $\mathbf{\cos^2 A + \sin^2 A = 1} \quad \implies \quad \mathbf{\sin^2 \theta + \cos^2 \theta = 1}$

• Domain: This identity is valid for all angles $\theta$ such that $0^\circ \le \theta \le 90^\circ$.
• Useful Algebraic Variations: $\sin^2 \theta = 1 - \cos^2 \theta \implies \sin \theta = \sqrt{1 - \cos^2 \theta}$ $\cos^2 \theta = 1 - \sin^2 \theta \implies \cos \theta = \sqrt{1 - \sin^2 \theta}$

Derivation 2: The Second Identity ($1 + an^2 heta = \sec^2 heta$)

Divide each term of the Pythagoras equation by $AB^2$ (the square of the base): $\frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$ $1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$

Since $\frac{BC}{AB} = \tan A$ and $\frac{AC}{AB} = \sec A$: $\mathbf{1 + \tan^2 \theta = \sec^2 \theta}$

• Domain: Valid for $0^\circ \le \theta < 90^\circ$ (at $\theta = 90^\circ$, $\tan 90^\circ$ and $\sec 90^\circ$ are undefined).
• Useful Algebraic Variations: $\sec^2 \theta - \tan^2 \theta = 1$ $\tan^2 \theta = \sec^2 \theta - 1$

The Reciprocal Difference-of-Squares Shortcut (CBSE High-Frequency Trick):

Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$: $\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$

$\mathbf{\sec \theta - \tan \theta = \frac{1}{\sec \theta + \tan \theta}}$ If a problem gives $\sec \theta + \tan \theta = 7$, then automatically $\sec \theta - \tan \theta = \frac{1}{7}$! Adding both equations solves for $\sec \theta$ in seconds.

Derivation 3: The Third Identity ($1 + \cot^2 heta = \csc^2 heta$)

Divide each term of the Pythagoras equation by $BC^2$ (the square of the perpendicular): $\frac{AB^2}{BC^2} + \frac{BC^2}{BC^2} = \frac{AC^2}{BC^2}$ $\left(\frac{AB}{BC}\right)^2 + 1 = \left(\frac{AC}{BC}\right)^2$

Since $\frac{AB}{BC} = \cot A$ and $\frac{AC}{BC} = \csc A$: $\mathbf{1 + \cot^2 \theta = \csc^2 \theta}$

• Domain: Valid for $0^\circ < \theta \le 90^\circ$ (at $\theta = 0^\circ$, $\cot 0^\circ$ and $\csc 0^\circ$ are undefined).
• Useful Algebraic Variations: $\csc^2 \theta - \cot^2 \theta = 1$ $\cot^2 \theta = \csc^2 \theta - 1$ $\mathbf{\csc \theta - \cot \theta = \frac{1}{\csc \theta + \cot \theta}}$

2. Expressing Ratios in Terms of a Single Function

In board examinations, you may be asked: "Express all trigonometric ratios in terms of $\sec A$".

• Cosine: $\cos A = \frac{1}{\sec A}$
• Sine: $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$
• Tangent: $\tan A = \sqrt{\sec^2 A - 1}$
• Cosecant: $\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$
• Cotangent: $\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$

3. Solved CBSE Board Examination Problems

Solved Example: The Difference of Squares Shortcut

Problem: If $\sec \theta + \tan \theta = p$, find the value of $\sec \theta$ and $\tan \theta$ in terms of $p$. Hence find $\sin \theta$.

Solution:

1. We are given: $\sec \theta + \tan \theta = p \quad \text{--- (1)}$
2. We know the identity $\sec^2 \theta - \tan^2 \theta = 1 \implies (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1$. Therefore: $\sec \theta - \tan \theta = \frac{1}{p} \quad \text{--- (2)}$
3. Add Equation (1) and Equation (2): $(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = p + \frac{1}{p}$ $2\sec \theta = \frac{p^2 + 1}{p} \implies \mathbf{\sec \theta = \frac{p^2 + 1}{2p}}$
4. Subtract Equation (2) from Equation (1): $(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = p - \frac{1}{p}$ $2\tan \theta = \frac{p^2 - 1}{p} \implies \mathbf{\tan \theta = \frac{p^2 - 1}{2p}}$
5. Find $\sin \theta$ using quotient relation: $\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{\frac{p^2 - 1}{2p}}{\frac{p^2 + 1}{2p}} = \mathbf{\frac{p^2 - 1}{p^2 + 1}}$

4. Summary and Examination Tips

Exam Tip: Whenever you see $1 - \cos^2 \theta$ or $1 - \sin^2 \theta$, immediately substitute $\sin^2 \theta$ or $\cos^2 \theta$. This simple substitution collapses multi-tier fractions instantly!

Common Mistake: Writing $\sec^2 \theta + \tan^2 \theta = 1$. The plus sign applies ONLY to sine and cosine ($\sin^2 \theta + \cos^2 \theta = 1$). For secant and tangent, the formula has a minus sign: $\sec^2 \theta - \tan^2 \theta = 1$!