Class 10 Chapter 8: Introduction to Trigonometry

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.

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 $B$ .

       A
       |\
       | \
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.

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:

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

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

Mnemonic Trick to Remember T-Ratios

A classic phrase popular in Indian classrooms:

\mathbf{P}\text{andit} & \mathbf{B}\text{adri} & \mathbf{P}\text{rasad} \\ \hline \mathbf{H}\text{ar} & \mathbf{H}\text{ar} & \mathbf{B}\text{ole} \end{array}$$ * $P / H = \sin$ * $B / H = \cos$ * $P / B = \tan$ --- ## 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}$ --- ### 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: 1. Write numbers $0, 1, 2, 3, 4$ under $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$. 2. 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$. 3. 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$. 4. **Congratulations! You just created the $\sin$ row!** 5. Reverse the $\sin$ row to get the $\cos$ row. 6. Divide $\sin / \cos$ to get the $\tan$ row. --- ## 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: --- ### 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$$ --- ### Identity 2: $1 + \tan^2\theta = \sec^2\theta$ **Proof:** Divide $AB^2 + BC^2 = AC^2$ by $BC^2$ (the Base squared): $$\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$$ --- ### Identity 3: $1 + \cot^2\theta = \text{cosec}^2\theta$ **Proof:** Divide $AB^2 + BC^2 = AC^2$ by $AB^2$ (the Perpendicular squared): $$1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$$ $$1 + \cot^2\theta = \text{cosec}^2\theta$$ --- ## 5. Step-by-Step Solved Practice Questions Let's test our understanding with 3 board-standard questions. --- ### Question 1: Finding T-ratios from a given ratio **Problem:** In a right $\Delta ABC$, right-angled at $B$, if $\tan A = \frac{4}{3}$, find all the other trigonometric ratios of angle $A$. #### Solution: 1. **Understand the given information:** $$\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{4}{3}$$ Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive real number. 2. **Find the Hypotenuse ($AC$) using Pythagoras Theorem:** $$AC^2 = AB^2 + BC^2$$ $$AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$$ $$AC = \sqrt{25k^2} = 5k$$ 3. **Calculate all remaining T-ratios:** * $\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \mathbf{\frac{4}{5}}$ * $\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \mathbf{\frac{3}{5}}$ * $\text{cosec } A = \frac{1}{\sin A} = \mathbf{\frac{5}{4}}$ * $\sec A = \frac{1}{\cos A} = \mathbf{\frac{5}{3}}$ * $\cot A = \frac{1}{\tan A} = \mathbf{\frac{3}{4}}$ --- ### Question 2: Evaluating expressions using angle values **Problem:** Evaluate the following expression: $$\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$$ #### Solution: 1. **Substitute values from the standard table:** * $\sin 30^\circ = \frac{1}{2}$ * $\tan 45^\circ = 1$ * $\text{cosec } 60^\circ = \frac{2}{\sqrt{3}}$ * $\sec 30^\circ = \frac{2}{\sqrt{3}}$ * $\cos 60^\circ = \frac{1}{2}$ * $\cot 45^\circ = 1$ 2. **Substitute these values into the expression:** $$\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}}$$ 3. **Divide Numerator by Denominator:** $$\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}$$ 4. **Rationalize the denominator:** $$\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}}$$ --- ### Question 3: Proving a Trigonometric Identity (NCERT Classic) **Problem:** Prove that: $$\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!}$$ --- ## 🌟 Quick Summary & Golden Rules for Exams 1. **Always double-check your reference angle** before taking Perpendicular vs Base. 2. **Memorize the $\sin \theta$ row** of the standard angle table; you can generate all other rows from it! 3. When proving identities, **converting everything to $\sin\theta$ and $\cos\theta$** is almost always a safe and effective strategy. 4. 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! --- ## Common Student Mistakes to Avoid 1. **Sign Errors in Algebraic Calculations**: Mistakes in distributing negative signs across brackets or when transferring terms across the equals sign. 2. **Formula Misapplication**: Memorizing formulas without checking required units or conditions (e.g. using diameter instead of radius). 3. **Skipping Intermediate Steps**: Jumping directly to final numerical answers without showing step-by-step mathematical working, leading to partial credit loss in board exams. 4. **Incorrect Unit Conversions**: Forgetting to convert parameters into uniform SI units (e.g., cm to meters or minutes to seconds) before computing. --- ## Exam Preparation & Frequently Asked Questions (FAQ) ### Q1. How should I revise Introduction to Trigonometry for the Class 10 Mathematics examination? Focus on mastering core textbook definitions, practicing 3-4 numerical problems daily with pen and paper, and reviewing previous year CBSE/NCERT board exam questions. ### Q2. What are the key concepts that carry maximum marks in this chapter? Pay special attention to core definitions, step-by-step derivations, solved textbook examples, and practical real-world applications outlined in your NCERT curriculum. ### Q3. How can I avoid losing marks in long answer questions? Always structure your answers with clear subheadings, write step-by-step working for numerical problems, state given values clearly, and highlight your final answers with correct SI units.

Class 10 Chapter 8: Introduction to Trigonometry

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.

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 $B$ .

       A
       |\
       | \
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.

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:

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

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

Mnemonic Trick to Remember T-Ratios

A classic phrase popular in Indian classrooms:

\mathbf{P}\text{andit} & \mathbf{B}\text{adri} & \mathbf{P}\text{rasad} \\ \hline \mathbf{H}\text{ar} & \mathbf{H}\text{ar} & \mathbf{B}\text{ole} \end{array}$$ * $P / H = \sin$ * $B / H = \cos$ * $P / B = \tan$ --- ## 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}$ --- ### 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: 1. Write numbers $0, 1, 2, 3, 4$ under $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$. 2. 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$. 3. 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$. 4. **Congratulations! You just created the $\sin$ row!** 5. Reverse the $\sin$ row to get the $\cos$ row. 6. Divide $\sin / \cos$ to get the $\tan$ row. --- ## 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: --- ### 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$$ --- ### Identity 2: $1 + \tan^2\theta = \sec^2\theta$ **Proof:** Divide $AB^2 + BC^2 = AC^2$ by $BC^2$ (the Base squared): $$\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$$ --- ### Identity 3: $1 + \cot^2\theta = \text{cosec}^2\theta$ **Proof:** Divide $AB^2 + BC^2 = AC^2$ by $AB^2$ (the Perpendicular squared): $$1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$$ $$1 + \cot^2\theta = \text{cosec}^2\theta$$ --- ## 5. Step-by-Step Solved Practice Questions Let's test our understanding with 3 board-standard questions. --- ### Question 1: Finding T-ratios from a given ratio **Problem:** In a right $\Delta ABC$, right-angled at $B$, if $\tan A = \frac{4}{3}$, find all the other trigonometric ratios of angle $A$. #### Solution: 1. **Understand the given information:** $$\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{4}{3}$$ Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive real number. 2. **Find the Hypotenuse ($AC$) using Pythagoras Theorem:** $$AC^2 = AB^2 + BC^2$$ $$AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$$ $$AC = \sqrt{25k^2} = 5k$$ 3. **Calculate all remaining T-ratios:** * $\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \mathbf{\frac{4}{5}}$ * $\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \mathbf{\frac{3}{5}}$ * $\text{cosec } A = \frac{1}{\sin A} = \mathbf{\frac{5}{4}}$ * $\sec A = \frac{1}{\cos A} = \mathbf{\frac{5}{3}}$ * $\cot A = \frac{1}{\tan A} = \mathbf{\frac{3}{4}}$ --- ### Question 2: Evaluating expressions using angle values **Problem:** Evaluate the following expression: $$\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$$ #### Solution: 1. **Substitute values from the standard table:** * $\sin 30^\circ = \frac{1}{2}$ * $\tan 45^\circ = 1$ * $\text{cosec } 60^\circ = \frac{2}{\sqrt{3}}$ * $\sec 30^\circ = \frac{2}{\sqrt{3}}$ * $\cos 60^\circ = \frac{1}{2}$ * $\cot 45^\circ = 1$ 2. **Substitute these values into the expression:** $$\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}}$$ 3. **Divide Numerator by Denominator:** $$\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}$$ 4. **Rationalize the denominator:** $$\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}}$$ --- ### Question 3: Proving a Trigonometric Identity (NCERT Classic) **Problem:** Prove that: $$\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!}$$ --- ## 🌟 Quick Summary & Golden Rules for Exams 1. **Always double-check your reference angle** before taking Perpendicular vs Base. 2. **Memorize the $\sin \theta$ row** of the standard angle table; you can generate all other rows from it! 3. When proving identities, **converting everything to $\sin\theta$ and $\cos\theta$** is almost always a safe and effective strategy. 4. 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! --- ## Common Student Mistakes to Avoid 1. **Sign Errors in Algebraic Calculations**: Mistakes in distributing negative signs across brackets or when transferring terms across the equals sign. 2. **Formula Misapplication**: Memorizing formulas without checking required units or conditions (e.g. using diameter instead of radius). 3. **Skipping Intermediate Steps**: Jumping directly to final numerical answers without showing step-by-step mathematical working, leading to partial credit loss in board exams. 4. **Incorrect Unit Conversions**: Forgetting to convert parameters into uniform SI units (e.g., cm to meters or minutes to seconds) before computing. --- ## Exam Preparation & Frequently Asked Questions (FAQ) ### Q1. How should I revise Introduction to Trigonometry for the Class 10 Mathematics examination? Focus on mastering core textbook definitions, practicing 3-4 numerical problems daily with pen and paper, and reviewing previous year CBSE/NCERT board exam questions. ### Q2. What are the key concepts that carry maximum marks in this chapter? Pay special attention to core definitions, step-by-step derivations, solved textbook examples, and practical real-world applications outlined in your NCERT curriculum. ### Q3. How can I avoid losing marks in long answer questions? Always structure your answers with clear subheadings, write step-by-step working for numerical problems, state given values clearly, and highlight your final answers with correct SI units.

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

Class 10 Chapter 8: Introduction to Trigonometry

1. The Foundation: The Right-Angled Triangle

2. Trigonometric Ratios (T-Ratios)

The Reciprocal Ratios:

Super Useful Quotient Relations:

Mnemonic Trick to Remember T-Ratios

Related Study Notes

Triangles

Circles

Algebraic Expressions and Identities

NCERT Study Guide Directory

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

Class 10 Chapter 8: Introduction to Trigonometry

1. The Foundation: The Right-Angled Triangle

2. Trigonometric Ratios (T-Ratios)

The Reciprocal Ratios:

Super Useful Quotient Relations:

Mnemonic Trick to Remember T-Ratios

Related Study Notes

Triangles

Circles

Algebraic Expressions and Identities

NCERT Study Guide Directory