Understanding Trigonometric Ratios and Identities

Trigonometric ratios and identities provide the foundational relationships between angles and the lengths of sides in right-angled triangles, leading to essential identities that enable deeper analysis in both pure and applied mathematics.

Trigonometric Ratios as Functions of Angles

For a right-angled triangle with angle $A$, hypotenuse of length $h$, side adjacent of length $b$, and side opposite of length $a$, the trigonometric ratios are formally defined as:

Sine: $\sin A = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{a}{h}$

Cosine: $\cos A = \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = \dfrac{b}{h}$

Tangent: $\tan A = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = \dfrac{a}{b}$

Cotangent: $\cot A = \dfrac{\text{Adjacent Side}}{\text{Opposite Side}} = \dfrac{b}{a}$

Secant: $\sec A = \dfrac{\text{Hypotenuse}}{\text{Adjacent Side}} = \dfrac{h}{b}$

Cosecant: $\csc A = \dfrac{\text{Hypotenuse}}{\text{Opposite Side}} = \dfrac{h}{a}$

The above definitions are valid for all $A$ such that the denominators are nonzero, extending beyond the geometric context via the unit circle definition.

Expressions of Ratios Using Coordinates

For a point $P(x, y)$ on a circle of radius $r$ making an angle $\theta$ with the positive $x$-axis, the ratios can be written as:

$\sin \theta = \dfrac{y}{r}$, $\cos \theta = \dfrac{x}{r}$, and $\tan \theta = \dfrac{y}{x}$ where $r = \sqrt{x^2 + y^2}$

This framework permits the extension of trigonometric ratios to all quadrants and is central to JEE Main's treatment as well as the NCERT curriculum. Refer to Trigonometry Overview for quadrant-wise sign convention.

Trigonometric Ratios for Standard Angles

Exact values of trigonometric ratios at principal angles (such as $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$) are foundational. For example, $\sin 0^\circ = 0$, $\sin 30^\circ = \dfrac{1}{2}$, $\sin 45^\circ = \dfrac{1}{\sqrt{2}}$, $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, and $\sin 90^\circ = 1$.

Parallel results apply for $\cos \theta$ and $\tan \theta$ at these standard angles. The knowledge of these exact values is critical for fast computations in competitive examinations. For a comprehensive formulae sheet, see Ratios and Identities.

Fundamental Trigonometric Identities: Statement and Proof

Identity 1: $\sin^2 \theta + \cos^2 \theta = 1$

Let $O$ be the origin, and $P$ be a point on the unit circle such that $OP = 1$ and $\angle XOP = \theta$, where $OX$ is the positive $x$-axis. By definition, the coordinates of $P$ are $(\cos \theta, \sin \theta)$. Thus, $OP^2 = (\cos \theta)^2 + (\sin \theta)^2 = 1^2$.

Therefore, $\sin^2 \theta + \cos^2 \theta = 1$ for all $\theta \in \mathbb{R}$.

Identity 2: $1 + \tan^2 \theta = \sec^2 \theta$

From the relation $\sin^2 \theta + \cos^2 \theta = 1$, divide both sides by $\cos^2 \theta$:

$\dfrac{\sin^2 \theta}{\cos^2 \theta} + \dfrac{\cos^2 \theta}{\cos^2 \theta} = \dfrac{1}{\cos^2 \theta}$

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

Thus, $1 + \tan^2 \theta = \sec^2 \theta$ holds for all $\theta \neq (2n+1)\dfrac{\pi}{2}$, $n \in \mathbb{Z}$.

Identity 3: $1 + \cot^2 \theta = \csc^2 \theta$

Again, from $\sin^2 \theta + \cos^2 \theta = 1$, divide by $\sin^2 \theta$:

$\dfrac{\sin^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{1}{\sin^2 \theta}$

$1 + \cot^2 \theta = \csc^2 \theta$

This identity is valid for all $\theta \neq n\pi$, $n \in \mathbb{Z}$ where $\sin \theta = 0$ is excluded from the domain.

Relation Between Ratios of Complementary Angles

For $0 < \theta < 90^\circ$,

$\sin(90^\circ-\theta) = \cos\theta$

$\cos(90^\circ-\theta) = \sin\theta$

$\tan(90^\circ-\theta) = \cot\theta$

$\cot(90^\circ-\theta) = \tan\theta$

$\sec(90^\circ-\theta) = \csc\theta$ and $\csc(90^\circ-\theta) = \sec\theta$

Reduction and Transformation Identities

Reduction identities relate trigonometric ratios of an angle to those of its reference angle. For any angle $\theta$:

$\sin(-\theta) = -\sin\theta$

$\cos(-\theta) = \cos\theta$

$\tan(-\theta) = -\tan\theta$

For angles $180^\circ\pm\theta$ and $360^\circ\pm\theta$:

$\sin(180^\circ - \theta) = \sin\theta$

$\cos(180^\circ - \theta) = -\cos\theta$

Such transformations are critical when simplifying trigonometric expressions. Application examples can be found at Trigonometric Ratios of Compound Angles.

Worked Example — Evaluation of Trigonometric Expression

Given: Compute $\sin^2 30^\circ + \cos^2 30^\circ$.

Substitution: $\sin 30^\circ = \dfrac{1}{2}$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$

Simplification: $\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{1}{4} + \dfrac{3}{4} = 1$

Final result: $\sin^2 30^\circ + \cos^2 30^\circ = 1$.

Complete Formula Collection: Trigonometric Ratios and Identities

For quick reference, the important formulas for trigonometric ratios and identities in JEE Main are:

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

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

$1 + \cot^2 \theta = \csc^2 \theta$

$\tan \theta = \dfrac{\sin \theta}{\cos \theta}$

$\cot \theta = \dfrac{\cos \theta}{\sin \theta}$

$\sin(90^\circ-\theta) = \cos\theta$

$\cos(90^\circ-\theta) = \sin\theta$

$\sin(-\theta) = -\sin\theta$

$\cos(-\theta) = \cos\theta$

For the complete set of formulas and quick review charts, visit Ratios and Identities.

Application in JEE Main and Beyond

Mastery of trigonometric ratios and identities is essential for solving problems in the JEE Main syllabus, especially those requiring transformation, simplification, and equation solving. These core results interlink with advanced topics such as inverse trigonometric functions, further discussed at Inverse Trigonometric Functions.