Trigonometry Notes

8

Introduction to Trigonometry

Class 10 Mathematics • Complete Board Exam Notes

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Chapter Roadmap

1. Right-Angled Triangle Basics 2. Trigonometric Ratios 3. Reciprocal Ratios 4. Standard Trigonometric Values 5. Complementary Angles 6. Trigonometric Identities 7. Finding Unknown Ratios & Angles 8. Board-Exam Questions 8.5 Ten Detailed Solved Questions 9. Exam Strategy 10. Quick Revision

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Chapter Goal

Trigonometry studies the relationship between the angles and sides of a triangle. In Class 10, the main focus is on a right-angled triangle, six trigonometric ratios, their standard values, complementary-angle relations and three basic identities.

1. Right-Angled Triangle Basics

1.1 The Three Important Sides

Consider a right-angled triangle with an acute angle \(\theta\). The side names depend on the angle \(\theta\) that we are considering.

• Hypotenuse (H): the side opposite the right angle. It is always the longest side.
• Perpendicular (P): the side opposite the angle \(\theta\).
• Base (B): the side adjacent to \(\theta\), other than the hypotenuse.

Important: The hypotenuse does not change when the reference angle changes, but the perpendicular and base can change because they are defined relative to \(\theta\).

1.2 Pythagoras Theorem

\[ \boxed{H^2=P^2+B^2} \]

This theorem is useful when one side of a right-angled triangle is unknown and the other two sides are known.

Solved Example 1 — Find the Hypotenuse

A right-angled triangle has perpendicular \(6\) cm and base \(8\) cm. Find its hypotenuse.

1. Use Pythagoras theorem: \[ H^2=P^2+B^2. \]
2. Substitute: \[ H^2=6^2+8^2=36+64=100. \]
3. Take the positive square root: \[ H=\sqrt{100}=10. \]

Answer: \(10\) cm

Solved Example 2 — Find the Perpendicular

The hypotenuse of a right triangle is \(13\) cm and the base is \(5\) cm. Find the perpendicular.

1. Use: \[ P^2=H^2-B^2. \]
2. Substitute: \[ P^2=13^2-5^2=169-25=144. \]
3. Therefore: \[ P=\sqrt{144}=12. \]

Answer: \(12\) cm

Solved Example 3 — Identify the Sides

For an acute angle \(\theta\) in a right-angled triangle, the side opposite \(\theta\) is \(9\) cm and the hypotenuse is \(15\) cm. What is the adjacent side?

1. The opposite side is \(P=9\) cm and the hypotenuse is \(H=15\) cm.
2. Use: \[ B^2=H^2-P^2. \]
3. Substitute: \[ B^2=15^2-9^2=225-81=144. \]
4. Hence: \[ B=12. \]

Answer: Adjacent side \(=12\) cm

2. Trigonometric Ratios

2.1 Six Ratios

For an acute angle \(\theta\) in a right-angled triangle, the six trigonometric ratios are defined as follows:

\[ \boxed{\sin\theta=\frac{P}{H}} \qquad \boxed{\cos\theta=\frac{B}{H}} \qquad \boxed{\tan\theta=\frac{P}{B}} \] \[ \boxed{\operatorname{cosec}\theta=\frac{H}{P}} \qquad \boxed{\sec\theta=\frac{H}{B}} \qquad \boxed{\cot\theta=\frac{B}{P}} \]

Memory trick: sin = P/H, cos = B/H, tan = P/B. The other three are their reciprocals.

2.2 Choosing the Correct Ratio

Read the question first. Then identify which two sides are involved. For example:

• Opposite and hypotenuse → \(\sin\theta\)
• Adjacent and hypotenuse → \(\cos\theta\)
• Opposite and adjacent → \(\tan\theta\)

Solved Example 1 — Find a Ratio

In a right triangle, \(P=3\) cm and \(H=5\) cm. Find \(\sin\theta\).

1. Use: \[ \sin\theta=\frac{P}{H}. \]
2. Substitute: \[ \sin\theta=\frac35. \]

Answer: \(\sin\theta=\frac35\)

Solved Example 2 — Find All Six Ratios

A right triangle has sides \(P=5\), \(B=12\), \(H=13\). Find all six trigonometric ratios of \(\theta\).

1. Use the definitions: \[ \sin\theta=\frac{P}{H}=\frac5{13}, \qquad \cos\theta=\frac{B}{H}=\frac{12}{13}. \]
2. Next: \[ \tan\theta=\frac{P}{B}=\frac5{12}. \]
3. Take reciprocals: \[ \operatorname{cosec}\theta=\frac{13}{5}, \qquad \sec\theta=\frac{13}{12}, \qquad \cot\theta=\frac{12}{5}. \]

Answer: \(\sin\theta=\frac5{13},\ \cos\theta=\frac{12}{13},\ \tan\theta=\frac5{12},\ \operatorname{cosec}\theta=\frac{13}{5},\ \sec\theta=\frac{13}{12},\ \cot\theta=\frac{12}{5}\)

Solved Example 3 — Find a Side from a Ratio

If \(\tan\theta=\frac34\) and the adjacent side is \(20\) cm, find the opposite side.

1. Use: \[ \tan\theta=\frac{P}{B}. \]
2. Substitute: \[ \frac34=\frac{P}{20}. \]
3. Cross multiply: \[ 4P=3(20)=60. \]
4. Therefore: \[ P=15. \]

Answer: Opposite side \(=15\) cm

3. Reciprocal Ratios

3.1 Reciprocal Relationships

The reciprocal of a non-zero number \(a\) is \(\frac1a\). Therefore:

\[ \boxed{\operatorname{cosec}\theta=\frac1{\sin\theta}} \qquad \boxed{\sec\theta=\frac1{\cos\theta}} \qquad \boxed{\cot\theta=\frac1{\tan\theta}} \]

Similarly:

\[ \boxed{\sin\theta=\frac1{\operatorname{cosec}\theta}} \qquad \boxed{\cos\theta=\frac1{\sec\theta}} \qquad \boxed{\tan\theta=\frac1{\cot\theta}} \]

Solved Example 1 — Reciprocal of Sine

If \(\sin\theta=\frac{4}{7}\), find \(\operatorname{cosec}\theta\).

1. Use: \[ \operatorname{cosec}\theta=\frac1{\sin\theta}. \]
2. Substitute: \[ \operatorname{cosec}\theta=\frac1{4/7}=\frac74. \]

Answer: \(\operatorname{cosec}\theta=\frac74\)

Solved Example 2 — Reciprocal of Cosine

If \(\cos\theta=\frac5{13}\), find \(\sec\theta\).

1. Use: \[ \sec\theta=\frac1{\cos\theta}. \]
2. Therefore: \[ \sec\theta=\frac1{5/13}=\frac{13}{5}. \]

Answer: \(\sec\theta=\frac{13}{5}\)

Solved Example 3 — Find the Original Ratio

If \(\cot\theta=\frac{9}{4}\), find \(\tan\theta\).

1. Use the reciprocal relation: \[ \tan\theta=\frac1{\cot\theta}. \]
2. Substitute: \[ \tan\theta=\frac1{9/4}=\frac49. \]

Answer: \(\tan\theta=\frac49\)

4. Standard Trigonometric Values

4.1 Complete Values of All Six Ratios

These standard values are extremely important for board exams. Learn the complete table carefully. The values of all six trigonometric ratios for \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\) are given below.

Angle	\(\sin\theta\)	\(\cos\theta\)	\(\tan\theta\)	\(\operatorname{cosec}\theta\)	\(\sec\theta\)	\(\cot\theta\)
\(0^\circ\)	\(0\)	\(1\)	\(0\)	Not defined	\(1\)	Not defined
\(30^\circ\)	\(\frac12\)	\(\frac{\sqrt3}{2}\)	\(\frac1{\sqrt3}\)	\(2\)	\(\frac2{\sqrt3}\)	\(\sqrt3\)
\(45^\circ\)	\(\frac1{\sqrt2}\)	\(\frac1{\sqrt2}\)	\(1\)	\(\sqrt2\)	\(\sqrt2\)	\(1\)
\(60^\circ\)	\(\frac{\sqrt3}{2}\)	\(\frac12\)	\(\sqrt3\)	\(\frac2{\sqrt3}\)	\(2\)	\(\frac1{\sqrt3}\)
\(90^\circ\)	\(1\)	\(0\)	Not defined	\(1\)	Not defined	\(0\)

One-table memory: \[ \sin\theta:\quad 0,\ \frac12,\ \frac1{\sqrt2},\ \frac{\sqrt3}{2},\ 1 \] \[ \cos\theta:\quad 1,\ \frac{\sqrt3}{2},\ \frac1{\sqrt2},\ \frac12,\ 0 \] \[ \tan\theta:\quad 0,\ \frac1{\sqrt3},\ 1,\ \sqrt3,\ \text{Not defined} \] The remaining three ratios are the reciprocals of these.

4.2 Why Some Values Are Not Defined

A fraction is not defined when its denominator is zero. This explains the special cases:

• \(\tan90^\circ=\frac{\sin90^\circ}{\cos90^\circ}=\frac10\), so it is not defined.
• \(\sec90^\circ=\frac1{\cos90^\circ}=\frac10\), so it is not defined.
• \(\cot0^\circ=\frac{\cos0^\circ}{\sin0^\circ}=\frac10\), so it is not defined.
• \(\operatorname{cosec}0^\circ=\frac1{\sin0^\circ}=\frac10\), so it is not defined.

4.3 Fast Memory Method

For sine, write the square roots of \(0,1,2,3,4\) over \(2\):

\[ \sin0^\circ=\frac{\sqrt0}{2}=0,\quad \sin30^\circ=\frac{\sqrt1}{2}=\frac12,\quad \sin45^\circ=\frac{\sqrt2}{2}=\frac1{\sqrt2}, \] \[ \sin60^\circ=\frac{\sqrt3}{2},\quad \sin90^\circ=\frac{\sqrt4}{2}=1. \]

The cosine values are the same sequence in reverse order. Then use reciprocal relationships for \(\operatorname{cosec},\sec,\cot\).

Solved Example 1 — Standard Value

Find \(\sin30^\circ+\cos60^\circ\).

1. Use standard values: \[ \sin30^\circ=\frac12,\qquad \cos60^\circ=\frac12. \]
2. Add: \[ \frac12+\frac12=1. \]

Answer: \(1\)

Solved Example 2 — Mixed Standard Values

Find \(2\sin60^\circ-\cos60^\circ\).

1. Substitute: \[ 2\left(\frac{\sqrt3}{2}\right)-\frac12. \]
2. Simplify: \[ \sqrt3-\frac12. \]

Answer: \(\sqrt3-\frac12\)

Solved Example 3 — Expression Using Standard Values

Find \(\tan45^\circ+\sin90^\circ-\cos0^\circ\).

1. Use: \[ \tan45^\circ=1,\quad \sin90^\circ=1,\quad \cos0^\circ=1. \]
2. Therefore: \[ 1+1-1=1. \]

Answer: \(1\)

4.4 All Six Ratios — Quick Check

\[ \operatorname{cosec}\theta=\frac1{\sin\theta},\qquad \sec\theta=\frac1{\cos\theta},\qquad \cot\theta=\frac1{\tan\theta}. \]

Exam check: At \(45^\circ\), \(\sin45^\circ=\cos45^\circ\) and \(\tan45^\circ=1\). At \(30^\circ\), \(\sin30^\circ=\cos60^\circ=\frac12\). At \(60^\circ\), \(\cos60^\circ=\sin30^\circ=\frac12\).

5. Complementary Angles

5.1 Meaning

Two angles are called complementary if their sum is \(90^\circ\).

\[ \boxed{A+B=90^\circ} \]

5.2 Complementary-Angle Relations

\[ \boxed{\sin(90^\circ-\theta)=\cos\theta} \] \[ \boxed{\cos(90^\circ-\theta)=\sin\theta} \] \[ \boxed{\tan(90^\circ-\theta)=\cot\theta} \] \[ \boxed{\cot(90^\circ-\theta)=\tan\theta} \] \[ \boxed{\sec(90^\circ-\theta)=\operatorname{cosec}\theta} \] \[ \boxed{\operatorname{cosec}(90^\circ-\theta)=\sec\theta} \]

Solved Example 1 — Convert to a Known Ratio

Find \(\sin(90^\circ-\theta)\) if \(\cos\theta=\frac35\).

1. Use: \[ \sin(90^\circ-\theta)=\cos\theta. \]
2. Given: \[ \cos\theta=\frac35. \]

Answer: \(\frac35\)

Solved Example 2 — Complementary Standard Angles

Find \(\sin30^\circ\) using a complementary-angle relation.

1. Since \(30^\circ=90^\circ-60^\circ\), \[ \sin30^\circ=\sin(90^\circ-60^\circ). \]
2. Use: \[ \sin(90^\circ-\theta)=\cos\theta. \]
3. Therefore: \[ \sin30^\circ=\cos60^\circ=\frac12. \]

Answer: \(\frac12\)

Solved Example 3 — Find a Complementary Ratio

If \(\tan\theta=\frac27\), find \(\cot(90^\circ-\theta)\).

1. Use: \[ \cot(90^\circ-\theta)=\tan\theta. \]
2. Given: \[ \tan\theta=\frac27. \]

Answer: \(\frac27\)

6. Trigonometric Identities

6.1 Meaning of an Identity

A trigonometric identity is an equation that is true for every value of the angle for which both sides are defined.

6.2 Three Basic Identities

\[ \boxed{\sin^2\theta+\cos^2\theta=1} \] \[ \boxed{1+\tan^2\theta=\sec^2\theta} \] \[ \boxed{1+\cot^2\theta=\operatorname{cosec}^2\theta} \]

Derived forms: \[ \sin^2\theta=1-\cos^2\theta,\qquad \cos^2\theta=1-\sin^2\theta \] \[ \tan^2\theta=\sec^2\theta-1,\qquad \cot^2\theta=\operatorname{cosec}^2\theta-1. \]

6.3 How to Prove an Identity

• Start with the side that looks more complicated.
• Convert \(\tan,\cot,\sec,\operatorname{cosec}\) into \(\sin,\cos\) when useful.
• Use a standard identity.
• Simplify until the other side is obtained.
• Do not start by assuming the identity you are trying to prove.

Solved Example 1 — Find Cosine from Sine

If \(\sin\theta=\frac35\), find \(\cos\theta\) for an acute angle \(\theta\).

1. Use: \[ \sin^2\theta+\cos^2\theta=1. \]
2. Substitute: \[ \left(\frac35\right)^2+\cos^2\theta=1. \]
3. Therefore: \[ \frac9{25}+\cos^2\theta=1 \Rightarrow\cos^2\theta=\frac{16}{25}. \]
4. Since \(\theta\) is acute, cosine is positive: \[ \cos\theta=\frac45. \]

Answer: \(\cos\theta=\frac45\)

Solved Example 2 — Prove an Identity

Prove that \(\displaystyle \frac{1-\cos^2\theta}{\sin\theta}=\sin\theta\).

1. Start with the left side: \[ \frac{1-\cos^2\theta}{\sin\theta}. \]
2. From \[ \sin^2\theta+\cos^2\theta=1, \] we get: \[ 1-\cos^2\theta=\sin^2\theta. \]
3. Therefore: \[ \frac{\sin^2\theta}{\sin\theta}=\sin\theta. \]

Hence proved.

Solved Example 3 — Prove Using a Reciprocal Identity

Prove that \(\displaystyle \frac{\sec^2\theta-1}{\tan\theta}=\tan\theta\).

1. Start with: \[ \frac{\sec^2\theta-1}{\tan\theta}. \]
2. Use: \[ \sec^2\theta-1=\tan^2\theta. \]
3. So: \[ \frac{\tan^2\theta}{\tan\theta}=\tan\theta. \]

Hence proved.

7. Finding Unknown Ratios & Angles

7.1 Finding an Unknown Ratio

When one trigonometric ratio is given, use an identity or the right-triangle relationship to find another ratio.

Solved Example 1 — Given Tangent

If \(\tan\theta=\frac34\), find \(\sec\theta\).

1. Use: \[ 1+\tan^2\theta=\sec^2\theta. \]
2. Substitute: \[ \sec^2\theta=1+\left(\frac34\right)^2 =1+\frac9{16} =\frac{25}{16}. \]
3. Since \(\theta\) is acute: \[ \sec\theta=\frac54. \]

Answer: \(\sec\theta=\frac54\)

7.2 Finding a Standard Angle

Solved Example 2 — Identify the Angle

If \(\sin\theta=\frac{\sqrt3}{2}\) and \(\theta\) is acute, find \(\theta\).

1. Recall the standard sine values.
2. \[ \sin60^\circ=\frac{\sqrt3}{2}. \]
3. Therefore: \[ \theta=60^\circ. \]

Answer: \(60^\circ\)

7.3 Simplifying a Trigonometric Expression

Solved Example 3 — Simplify

Simplify \(\displaystyle \frac{\sin\theta}{\cos\theta}\).

1. Use the definition: \[ \tan\theta=\frac{\sin\theta}{\cos\theta}. \]
2. Therefore: \[ \frac{\sin\theta}{\cos\theta}=\tan\theta. \]

Answer: \(\tan\theta\)

8. Board-Exam Questions

Board Pattern 1

Question: If \(\sin\theta=\frac{12}{13}\), find \(\cos\theta\) and \(\tan\theta\), where \(\theta\) is acute.

Solution:

\[ \sin^2\theta+\cos^2\theta=1 \] \[ \cos^2\theta=1-\frac{144}{169} =\frac{25}{169} \] \[ \cos\theta=\frac5{13}. \] Then \[ \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{12/13}{5/13} =\frac{12}{5}. \]

Answer: \(\cos\theta=\frac5{13},\ \tan\theta=\frac{12}{5}\)

Board Pattern 2

Question: Evaluate \(\sin^230^\circ+\cos^260^\circ\).

Solution:

\[ \sin30^\circ=\frac12,\qquad \cos60^\circ=\frac12. \] \[ \sin^230^\circ+\cos^260^\circ =\left(\frac12\right)^2+\left(\frac12\right)^2 =\frac14+\frac14 =\frac12. \]

Answer: \(\frac12\)

Board Pattern 3

Question: Prove that \(\displaystyle \frac{1-\sin^2\theta}{\cos\theta}=\cos\theta\).

Solution:

\[ \frac{1-\sin^2\theta}{\cos\theta} \] \[ =\frac{\cos^2\theta}{\cos\theta} =\cos\theta. \]

Hence proved.

Board Pattern 4

Question: If \(\tan\theta=\frac1{\sqrt3}\), where \(\theta\) is acute, find \(\theta\).

Solution:

\[ \tan30^\circ=\frac1{\sqrt3}. \] Therefore: \[ \theta=30^\circ. \]

Answer: \(30^\circ\)

8.1 High-Scoring Presentation Format

1. Write the correct trigonometric ratio first.
2. Identify \(P,B,H\) relative to the given angle.
3. Substitute values carefully.
4. For identities, work from one side only.
5. Use standard values instead of decimal approximations.
6. Keep radicals in exact form, such as \(\sqrt3\), unless a decimal is specifically requested.
7. End with a clear final answer.

8.5 Ten Detailed Solved Questions

Question 1

If \(\sin\theta=\frac35\), find \(\cos\theta,\tan\theta,\sec\theta,\operatorname{cosec}\theta,\cot\theta\), where \(\theta\) is acute.

1. Start with: \[ \sin\theta=\frac35. \]
2. Use: \[ \sin^2\theta+\cos^2\theta=1. \] Therefore, \[ \cos^2\theta=1-\frac9{25}=\frac{16}{25}. \]
3. Since \(\theta\) is acute: \[ \cos\theta=\frac45. \]
4. Now: \[ \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{3/5}{4/5}=\frac34. \]
5. Take reciprocals: \[ \sec\theta=\frac54,\qquad \operatorname{cosec}\theta=\frac53,\qquad \cot\theta=\frac43. \]

Answer: \(\cos\theta=\frac45,\ \tan\theta=\frac34,\ \sec\theta=\frac54,\ \operatorname{cosec}\theta=\frac53,\ \cot\theta=\frac43\)

Question 2

If \(\tan\theta=\frac5{12}\), find \(\sin\theta\) and \(\cos\theta\), where \(\theta\) is acute.

1. From \[ \tan\theta=\frac{P}{B}=\frac5{12}, \] take \(P=5\) and \(B=12\).
2. Find the hypotenuse: \[ H=\sqrt{P^2+B^2} =\sqrt{5^2+12^2} =\sqrt{25+144}=13. \]
3. Therefore: \[ \sin\theta=\frac{P}{H}=\frac5{13}. \]
4. Also: \[ \cos\theta=\frac{B}{H}=\frac{12}{13}. \]

Answer: \(\sin\theta=\frac5{13},\ \cos\theta=\frac{12}{13}\)

Question 3

Evaluate \(\sin^230^\circ+\cos^260^\circ+\tan^245^\circ\).

1. Use: \[ \sin30^\circ=\frac12,\quad \cos60^\circ=\frac12,\quad \tan45^\circ=1. \]
2. Square the values: \[ \sin^230^\circ=\frac14,\quad \cos^260^\circ=\frac14,\quad \tan^245^\circ=1. \]
3. Add: \[ \frac14+\frac14+1=\frac12+1=\frac32. \]

Answer: \(\frac32\)

Question 4

Prove that \(\displaystyle \frac{1-\sin^2\theta}{\cos\theta}=\cos\theta\).

1. Start from the left-hand side: \[ \frac{1-\sin^2\theta}{\cos\theta}. \]
2. Use: \[ \sin^2\theta+\cos^2\theta=1. \]
3. Hence: \[ 1-\sin^2\theta=\cos^2\theta. \]
4. Substitute: \[ \frac{\cos^2\theta}{\cos\theta}=\cos\theta. \]

Hence proved.

Question 5

If \(\sec\theta=\frac{13}{12}\), find \(\tan\theta\), where \(\theta\) is acute.

1. Use: \[ 1+\tan^2\theta=\sec^2\theta. \]
2. Substitute: \[ 1+\tan^2\theta=\left(\frac{13}{12}\right)^2 =\frac{169}{144}. \]
3. Therefore: \[ \tan^2\theta=\frac{169}{144}-1 =\frac{25}{144}. \]
4. Since \(\theta\) is acute: \[ \tan\theta=\frac5{12}. \]

Answer: \(\tan\theta=\frac5{12}\)

Question 6

Evaluate \(\displaystyle \frac{\sin60^\circ}{\cos30^\circ}+\frac{\tan45^\circ}{\sec60^\circ}\).

1. Use standard values: \[ \sin60^\circ=\frac{\sqrt3}{2},\quad \cos30^\circ=\frac{\sqrt3}{2},\quad \tan45^\circ=1,\quad \sec60^\circ=2. \]
2. Substitute: \[ \frac{\sqrt3/2}{\sqrt3/2}+\frac12. \]
3. Simplify: \[ 1+\frac12=\frac32. \]

Answer: \(\frac32\)

Question 7

If \(\cot\theta=\frac7{24}\), find \(\sin\theta\) and \(\sec\theta\), where \(\theta\) is acute.

1. Since \[ \cot\theta=\frac{B}{P}=\frac7{24}, \] take \(B=7\) and \(P=24\).
2. Find \(H\): \[ H=\sqrt{7^2+24^2} =\sqrt{49+576} =25. \]
3. Therefore: \[ \sin\theta=\frac{P}{H}=\frac{24}{25}. \]
4. Also: \[ \sec\theta=\frac{H}{B}=\frac{25}{7}. \]

Answer: \(\sin\theta=\frac{24}{25},\ \sec\theta=\frac{25}{7}\)

Question 8

Prove that \(\displaystyle \frac{\sec^2\theta-1}{\tan^2\theta}=1\).

1. Start with: \[ \frac{\sec^2\theta-1}{\tan^2\theta}. \]
2. Use the identity: \[ \sec^2\theta=1+\tan^2\theta. \]
3. Therefore: \[ \sec^2\theta-1=\tan^2\theta. \]
4. So: \[ \frac{\tan^2\theta}{\tan^2\theta}=1. \]

Hence proved.

Question 9

If \(A+B=90^\circ\), prove that \(\sin A=\cos B\).

1. Given: \[ A+B=90^\circ. \]
2. Rearrange: \[ A=90^\circ-B. \]
3. Take sine: \[ \sin A=\sin(90^\circ-B). \]
4. Use the complementary-angle relation: \[ \sin(90^\circ-B)=\cos B. \]

Hence, \(\sin A=\cos B\).

Question 10

If \(\sin\theta=\frac8{17}\), find \(\displaystyle \frac{1+\cos\theta}{1-\cos\theta}\), where \(\theta\) is acute.

1. Use: \[ \sin^2\theta+\cos^2\theta=1. \]
2. Substitute: \[ \left(\frac8{17}\right)^2+\cos^2\theta=1. \]
3. Therefore: \[ \cos^2\theta=1-\frac{64}{289} =\frac{225}{289}. \]
4. Since \(\theta\) is acute: \[ \cos\theta=\frac{15}{17}. \]
5. Substitute: \[ \frac{1+\cos\theta}{1-\cos\theta} = \frac{1+\frac{15}{17}}{1-\frac{15}{17}}. \]
6. Simplify numerator and denominator: \[ = \frac{\frac{32}{17}}{\frac{2}{17}} =\frac{32}{2}=16. \]

Answer: \(16\)

9. Exam Strategy

1. Master the six definitions.

Remember: \(\sin=P/H,\ \cos=B/H,\ \tan=P/B\), and their reciprocals.

2. Learn the standard-value table.

The values of \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\) appear frequently in direct evaluation and identity questions.

3. Use identities strategically.

If \(\sin\theta\) is given, \(\sin^2\theta+\cos^2\theta=1\) is usually the quickest way to obtain \(\cos\theta\). If \(\tan\theta\) is given, use \(1+\tan^2\theta=\sec^2\theta\).

4. Watch undefined values.

\(\tan90^\circ\), \(\sec90^\circ\), \(\cot0^\circ\), and \(\operatorname{cosec}0^\circ\) are not defined because their denominator becomes zero.

Most Common Errors

• Confusing perpendicular and base for the chosen angle.
• Writing \(\sin=P/B\) or \(\tan=P/H\).
• Forgetting that reciprocal ratios are flipped fractions.
• Mixing up \(\sin30^\circ\) and \(\cos30^\circ\).
• Forgetting the square in \(\sin^2\theta\).
• Using an identity incorrectly while proving another identity.
• Giving a negative ratio for an acute angle.
• Using decimal approximations when exact values are expected.

10. Quick Revision Sheet

Concept	Formula / Key Fact
Right Triangle	\(H^2=P^2+B^2\)
Sine	\(\sin\theta=\frac{P}{H}\)
Cosine	\(\cos\theta=\frac{B}{H}\)
Tangent	\(\tan\theta=\frac{P}{B}\)
Cosecant	\(\operatorname{cosec}\theta=\frac{H}{P}\)
Secant	\(\sec\theta=\frac{H}{B}\)
Cotangent	\(\cot\theta=\frac{B}{P}\)
Identity 1	\(\sin^2\theta+\cos^2\theta=1\)
Identity 2	\(1+\tan^2\theta=\sec^2\theta\)
Identity 3	\(1+\cot^2\theta=\operatorname{cosec}^2\theta\)
Complementary	\(\sin(90^\circ-\theta)=\cos\theta\)
Standard Value	\(\sin30^\circ=\frac12,\ \cos60^\circ=\frac12,\ \tan45^\circ=1\)

30-Second Final Revision

• First identify: perpendicular, base and hypotenuse.
• Then choose: sin, cos or tan according to the two required sides.
• Remember: cosec, sec and cot are reciprocals.
• Learn: standard values of \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\).
• Use: the three fundamental identities for missing ratios and proofs.
• Board answer: formula → substitution → simplification → final statement.

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Class 10 Mathematics • Chapter 8: Introduction to Trigonometry

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Practise definitions, standard values and identities for maximum accuracy.