Introduction to Trigonometry - Trigonometric ratios, values for specific angles (30, 45, 60), and trigonometric identities

Welcome, standard 10 students! Today, we are diving into one of the most fascinating and high-scoring chapters in your Class 10 NCERT Mathematics syllabus: Introduction to Trigonometry.

Have you ever looked at a super tall building like the Qutub Minar and wondered, *"How did engineers measure its height without climbing all the way up with a giant tape measure?"*

The secret weapon they used is Trigonometry!

The word *Trigonometry* comes from three Greek words:

Tri = Three

Gon = Sides

Metron = Measure

In simple terms, Trigonometry is the study of relationships between the sides and angles of a triangle.

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1. The Foundation: The Right-Angled Triangle

Trigonometry in Class 10 revolves around Right-Angled Triangles.

Let’s take a right-angled triangle, $\Delta ABC$, right-angled at $B$.

```

| \

Opp/ | \ Hypotenuse (H)

Perp | \

(P) | \

|_____\

B (Base/Adj) C

Angle θ

```

To work with trigonometric ratios, you must identify three key sides relative to an acute angle (let's call it $\theta$ or Angle $C$):

Hypotenuse ($H$): The longest side, directly opposite the $90^\circ$ angle. (This never changes!)

Perpendicular / Opposite ($P$): The side directly opposite to the reference angle $\theta$.

Base / Adjacent ($B$): The side adjacent to (touching) angle $\theta$ (other than the hypotenuse).

⚠️ Teacher's Tip: Perpendicular and Base change depending on which angle you are looking at!

* If we look from Angle $C$: Side $AB$ is Perpendicular, Side $BC$ is Base.

* If we look from Angle $A$: Side $BC$ becomes Perpendicular, Side $AB$ becomes Base.

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2. Trigonometric Ratios (T-Ratios)

Trigonometric ratios are simple ratios of the lengths of two sides of a right-angled triangle with respect to its acute angles.

There are 6 fundamental ratios:

$$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{P}{H}$$

$$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{B}{H}$$

$$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{P}{B}$$

The Reciprocal Ratios:

Each of the primary ratios has a reciprocal pair:

Cosecant ($\text{cosec } \theta$): Reciprocal of $\sin \theta \implies \frac{H}{P} = \frac{1}{\sin \theta}$

Secant ($\sec \theta$): Reciprocal of $\cos \theta \implies \frac{H}{B} = \frac{1}{\cos \cos \theta}$

Cotangent ($\cot \theta$): Reciprocal of $\tan \theta \implies \frac{B}{P} = \frac{1}{\tan \theta}$

Super Useful Quotient Relations:

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

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

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Mnemonic Trick to Remember T-Ratios

A classic phrase popular in Indian classrooms:

$$\begin{array}{rcccl}

\text{\textbf{P}andit} & \text{\textbf{B}adri} & \text{\textbf{P}rasad} \\

\hline

\text{\textbf{H}ar} & \text{\textbf{H}ar} & \text{\textbf{B}ole}

\end{array}$$

$P / H = \sin$

$B / H = \cos$

$P / B = \tan$

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3. Trigonometric Values of Specific Angles ($30^\circ$, $45^\circ$, $60^\circ$)

While you can calculate ratios for any angle, your Class 10 board exam focuses heavily on specific standard angles: $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$.

Geometric Derivations (Understanding the "Why")

A. Ratios of $45^\circ$

Consider an isosceles right triangle where base = perpendicular = $a$.

By Pythagoras theorem: $\text{Hypotenuse} = \sqrt{a^2 + a^2} = a\sqrt{2}$

$\sin 45^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}} = \frac{a}{a} = 1$

B. Ratios of $30^\circ$ and $60^\circ$

Consider an equilateral triangle $ABC$ with side length $2a$. Drop a perpendicular $AD$ from $A$ to $BC$.

$BD = a$, $\angle BAD = 30^\circ$, $\angle ABD = 60^\circ$.

Using Pythagoras theorem in $\Delta ABD$: $AD = \sqrt{(2a)^2 - a^2} = a\sqrt{3}$.

$\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}$

$\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$

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Complete NCERT Table of Specific Angles

Here is the master table you need to memorize:

Ratio	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin \theta$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos \theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$
$\tan \theta$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	Not Defined
$\text{cosec } \theta$	Not Defined	$2$	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	$1$
$\sec \theta$	$1$	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	Not Defined
$\cot \theta$	Not Defined	$\sqrt{3}$	$1$	$\frac{1}{\sqrt{3}}$	$0$

💡 Shortcut Trick to Build the Table Yourself:

Write numbers $0, 1, 2, 3, 4$ under $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$.

Divide each number by $4$: $\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4} \rightarrow 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$.

Take the square root of each result: $\sqrt{0}=0, \sqrt{\frac{1}{4}}=\frac{1}{2}, \sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}, \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}, \sqrt{1}=1$.

Congratulations! You just created the $\sin$ row!

Reverse the $\sin$ row to get the $\cos$ row.

Divide $\sin / \cos$ to get the $\tan$ row.

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4. Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it holds true for all values of the angle(s) involved.

There are three fundamental trigonometric identities in Class 10:

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

Proof:

In a right-angled triangle $\Delta ABC$, by Pythagoras Theorem:

$$AB^2 + BC^2 = AC^2$$

Divide both sides by $AC^2$:

$$\left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = \left(\frac{AC}{AC}\right)^2$$

Since $\frac{AB}{AC} = \sin\theta$ and $\frac{BC}{AC} = \cos\theta$:

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

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

Proof:

Divide $AB^2 + BC^2 = AC^2$ by $BC^2$ (the Base squared):

$$\left(\frac{AB}{BC}\right)^2 + 1 = \left(\frac{AC}{BC}\right)^2$$

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

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Identity 3: $1 + \cot^2\theta = \text{cosec}^2\theta$

Proof:

Divide $AB^2 + BC^2 = AC^2$ by $AB^2$ (the Perpendicular squared):

$$1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$$

$$1 + \cot^2\theta = \text{cosec}^2\theta$$

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5. Step-by-Step Solved Practice Questions

Let's test our understanding with 3 board-standard questions.

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Question 1: Finding T-ratios from a given ratio

Problem: In a right $\Delta ABC$, right-angled at $B$, if $\tan A = \frac{4}{3}$, find all the other trigonometric ratios of angle $A$.

Solution:

Understand the given information:

$$\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{4}{3}$$

Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive real number.

Find the Hypotenuse ($AC$) using Pythagoras Theorem:

$$AC^2 = AB^2 + BC^2$$

$$AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$$

$$AC = \sqrt{25k^2} = 5k$$

Calculate all remaining T-ratios:

$\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \mathbf{\frac{4}{5}}$

$\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \mathbf{\frac{3}{5}}$

$\text{cosec } A = \frac{1}{\sin A} = \mathbf{\frac{5}{4}}$

$\sec A = \frac{1}{\cos A} = \mathbf{\frac{5}{3}}$

$\cot A = \frac{1}{\tan A} = \mathbf{\frac{3}{4}}$

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Question 2: Evaluating expressions using angle values

Problem: Evaluate the following expression:

$$\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$$

Solution:

Substitute values from the standard table:

$\sin 30^\circ = \frac{1}{2}$

$\tan 45^\circ = 1$

$\text{cosec } 60^\circ = \frac{2}{\sqrt{3}}$

$\sec 30^\circ = \frac{2}{\sqrt{3}}$

$\cos 60^\circ = \frac{1}{2}$

$\cot 45^\circ = 1$

Substitute these values into the expression:

$$\text{Numerator} = \frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}}$$

$$\text{Denominator} = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}}$$

Divide Numerator by Denominator:

$$\text{Expression} = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{3\sqrt{3} + 4}{2\sqrt{3}}} = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$$

Rationalize the denominator:

$$\frac{(3\sqrt{3} - 4)(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)} = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - (4)^2}$$

$$= \frac{(27 + 16 - 24\sqrt{3})}{27 - 16} = \mathbf{\frac{43 - 24\sqrt{3}}{11}}$$

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Question 3: Proving a Trigonometric Identity (NCERT Classic)

Problem: Prove that:

$$\frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta$$

Solution:

Step 1: Take the Left-Hand Side (LHS):

$$\text{LHS} = \frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta}$$

Step 2: Factor out common terms from numerator and denominator:

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

Step 3: Use the identity $\sin^2\theta = 1 - \cos^2\theta$ in the numerator:

$$1 - 2\sin^2\theta = 1 - 2(1 - \cos^2\theta)$$

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

Step 4: Substitute this back into the expression:

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

Step 5: Cancel out the identical term $(2\cos^2\theta - 1)$:

$$\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$$

$$\text{Hence Proved!}$$

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🌟 Quick Summary & Golden Rules for Exams

Always double-check your reference angle before taking Perpendicular vs Base.

Memorize the $\sin \theta$ row of the standard angle table; you can generate all other rows from it!

When proving identities, converting everything to $\sin\theta$ and $\cos\theta$ is almost always a safe and effective strategy.

Keep basic algebraic formulas handy:

$(a+b)^2 = a^2 + 2ab + b^2$

$a^2 - b^2 = (a-b)(a+b)$

Keep practicing, stay confident, and you will easily score full marks in Trigonometry!