SEBA Class 10 Maths Chapter 8.3 Introduction to Trigonometry | New Book

Get the Free SEBA Class 10 Maths Chapter 8.3 Introduction to Trigonometry (New Book) Solution. This article provides complete solutions to Exercise 8.3 in a simple way based on the new SEBA Class 10 Maths textbook. These Class 10 Maths Chapter 8.3 solutions will help you understand the basics of Introduction to Trigonometry and prepare for the upcoming HSLC examination. 

See More:

Chapter 8.1 Introduction to Trigonometry

Chapter 8.2 Introduction to Trigonometry

Chapter 8.3 Introduction to Trigonometry

Chapter 8.4 Introduction to Trigonometry

Key Formulas to Remember

Before starting, we use these complementary angle rules:

• $\sin(90^\circ – \theta) = \cos\theta$ and $\cos(90^\circ – \theta) = \sin\theta$
• $\tan(90^\circ – \theta) = \cot\theta$ and $\cot(90^\circ – \theta) = \tan\theta$
• $\sec(90^\circ – \theta) = cosec \theta$ and $cosec (90^\circ – \theta) = \sec\theta$
• $\tan\theta \times \cot\theta = 1$

1. Evaluate:

(i) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$

Sol.

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}} $

$= \frac{\sin(90^{\circ} – 72^{\circ})}{\cos 72^{\circ}}$ $\quad [\because \sin(90^{\circ} – \theta) = \cos \theta]$

$= \frac{\cos 72^{\circ}}{\cos 72^{\circ}}$

$= 1$

(ii) $\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$

Sol.

$\frac{\tan 26^{\circ}}{\cot 64^{\circ}} $

$= \frac{\tan(90^{\circ} – 64^{\circ})}{\cot 64^{\circ}}$ $\quad [\because \tan(90^{\circ} – \theta) = \cot\theta]$ 

$= \frac{\cot 64^{\circ}}{\cot 64^{\circ}}$

$= 1$

(iii) $\cos 48^{\circ} – \sin 42^{\circ}$

Sol.

$\cos 48^{\circ} – \sin 42^{\circ}$

$= \cos(90^{\circ} – 42^{\circ}) – \sin 42^{\circ}$ $\quad [\because \cos(90^{\circ} – \theta) = \sin \theta]$

$= \sin 42^{\circ} – \sin 42^{\circ}$

$= 0$

(iv) $cosec 31^{\circ} – \sec 59^{\circ}$

Sol.

$cosec 31^{\circ} – \sec 59^{\circ} $

$= cosec(90^{\circ} – 59^{\circ}) – \sec 59^{\circ}$ $\quad [\because cosec(90^{\circ} – \theta) = \sec \theta]$

$= \sec 59^{\circ} – \sec 59^{\circ}$

$= 0$

(v) $\sin 35^{\circ} \sin 55^{\circ} – \cos 35^{\circ} \cos 55^{\circ}$

Sol.

$\sin 35^{\circ} \sin 55^{\circ} – \cos 35^{\circ} \cos 55^{\circ}$

$= \sin(90^{\circ} – 55^{\circ}) \sin(90^{\circ} – 35^{\circ}) – \cos 35^{\circ} \cos 55^{\circ}$

$= \cos 55^{\circ} \cos 35^{\circ} – \cos 35^{\circ} \cos 55^{\circ}$

$= 0$

(vi) $\tan 35^{\circ} \tan 60^{\circ} \tan 55^{\circ} \tan 30^{\circ}$

Sol. $\tan 35^{\circ} \tan 60^{\circ} \tan 55^{\circ} \tan 30^{\circ}$

$= \tan(90^{\circ} – 55^{\circ}) \tan(90^{\circ} – 30^{\circ}) \cdot \tan 55^{\circ} \cdot \tan 30^{\circ}$ $\quad [\because \tan(90^{\circ} – A) = \cot A]$

$= \cot 55^{\circ} \cdot \cot 30^{\circ} \cdot \tan 55^{\circ} \cdot \tan 30^{\circ}$

$= (\cot 55^{\circ} \times \tan 55^{\circ}) (\cot 30^{\circ} \times \tan 30^{\circ})$

$= \left(\frac{1}{\tan 55^{\circ}} \times \tan 55^{\circ}\right) \left(\frac{1}{\tan 30^{\circ}} \times \tan 30^{\circ}\right)$ $\quad [\because \cot A = \frac{1}{\tan A}]$

$= 1 \times 1$

$= 1$

(vii) $\frac{\cot 54^{\circ}}{\tan 36^{\circ}} + \frac{\tan 20^{\circ}}{\cot 70^{\circ}} – 2$

Sol.

$\frac{\cot 54^{\circ}}{\tan 36^{\circ}} + \frac{\tan 20^{\circ}}{\cot 70^{\circ}} – 2$

$= \frac{\cot(90^{\circ} – 36^{\circ})}{\tan 36^{\circ}} + \frac{\tan(90^{\circ} – 70^{\circ})}{\cot 70^{\circ}} – 2$

$= \frac{\tan 36^{\circ}}{\tan 36^{\circ}} + \frac{\cot 70^{\circ}}{\cot 70^{\circ}} – 2$

$= 1 + 1 – 2$

$= 2 – 2$

$= 0$

(viii) $3\frac{\sin 23^{\circ}}{\cos 67^{\circ}} + 4\frac{\sec 47^{\circ}}{cosec 43^{\circ}}$

Sol.

$3\frac{\sin 23^{\circ}}{\cos 67^{\circ}} + 4\frac{\sec 47^{\circ}}{cosec 43^{\circ}}$

$= \frac{3 \sin(90^{\circ} – 67^{\circ})}{\cos 67^{\circ}} + \frac{4 \sec(90^{\circ} – 43^{\circ})}{cosec 43^{\circ}}$

$= \frac{3 \cos 67^{\circ}}{\cos 67^{\circ}} + \frac{4 \csc 43^{\circ}}{cosec 43^{\circ}}$

$= 3 + 4$

$= 7$

(ix) $\tan 5^{\circ} \tan 25^{\circ} \tan 30^{\circ} \tan 65^{\circ} \tan 85^{\circ}$

Sol.

$\tan 5^{\circ} \tan 25^{\circ} \tan 30^{\circ} \tan 65^{\circ} \tan 85^{\circ}$

$= \tan(90^{\circ} – 85^{\circ}) \tan(90^{\circ} – 65^{\circ}) \times \frac{1}{\sqrt{3}} \times \tan 65^{\circ} \tan 85^{\circ}$

$= \cot 85^{\circ} \cdot \cot 65^{\circ} \times \frac{1}{\sqrt{3}} \times \frac{1}{\cot 65^{\circ}} \times \frac{1}{\cot 85^{\circ}}$

$= \frac{1}{\sqrt{3}}$

2. Show that:

(i) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$

Sol.

LHS $= \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

$= (\tan 48^{\circ} \tan 42^{\circ})(\tan 23^{\circ} \tan 67^{\circ})$

$= \tan(90^{\circ} – 42^{\circ}) \tan 42^{\circ} \tan(90^{\circ} – 67^{\circ}) \tan 67^{\circ}$ $\quad [\because \tan(90^{\circ} – A) = \cot A]$

$= \cot 42^{\circ} \tan 42^{\circ} \cot 67^{\circ} \tan 67^{\circ}$

$= (\frac{1}{\tan 42^{\circ}} \times \tan 42^{\circ})(\frac{1}{\tan 67^{\circ}} \times \tan 67^{\circ})$ $\quad [\because \cot A = \frac{1}{\tan A}]$

$= 1 \times 1$

$= 1$

$= \text{RHS}$

$\text{LHS} = \text{RHS}$

∴ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$

Hence proved

(ii) $\cos 38^{\circ} \cos 52^{\circ} – \sin 38^{\circ} \sin 52^{\circ} = 0$

Sol.

LHS $= \cos 38^{\circ} \cos 52^{\circ} – \sin 38^{\circ} \sin 52^{\circ}$

$= \cos(90^{\circ} – 52^{\circ}) \cos(90^{\circ} – 38^{\circ}) – \sin 38^{\circ} \sin 52^{\circ}$ $\quad [\because \cos(90^{\circ} – A) = \sin A]$

$= \sin 52^{\circ} \sin 38^{\circ} – \sin 38^{\circ} \sin 52^{\circ}$

$= \sin 52^{\circ} \sin 38^{\circ} – \sin 52^{\circ} \sin 38^{\circ}$

$= 0$

$= \text{RHS}$

$\text{LHS} = \text{RHS}$

∴ $\cos 38^{\circ} \cos 52^{\circ} – \sin 38^{\circ} \sin 52^{\circ} = 0$

Hence proved

3. If $\tan 2A = \cot(A – 18^{\circ})$; where $2A$ is an acute angle, find the value of $A$.

Sol.

Given, $\tan 2A = \cot(A – 18^{\circ})$

$\Rightarrow \cot(90^{\circ} – 2A) = \cot(A – 18^{\circ})$ $\quad [\because \tan \theta = \cot(90^{\circ} – \theta)]$

$\Rightarrow 90^{\circ} – 2A = A – 18^{\circ}$ $\quad [\because (90^{\circ} – 2A) \text{ and } (A – 18^{\circ}) \text{ are both acute angles}]$

$\Rightarrow 90 – 2A = A – 18$

$\Rightarrow 2A + A = 90^{\circ} + 18^{\circ}$

$\Rightarrow 3A = 108^{\circ}$

$\Rightarrow A = \frac{108^{\circ}}{3}$

$\Rightarrow A = 36^{\circ}$

Therefore, the required value of $\angle A = 36^{\circ}$

4. If $\tan A = \cot B$, prove that $A + B = 90^{\circ}$.

Sol.

Given, $\tan A = \cot B$

$\Rightarrow \tan A = \tan(90^{\circ} – B)$

$\Rightarrow A = 90^{\circ} – B$ $\quad [\because A \text{ and } (90^{\circ} – B) \text{ are both acute angles}]$

$\Rightarrow A + B = 90^{\circ}$

Hence proved

5. If $\sec 4A = cosec(A – 20^{\circ})$ where $4A$ is an acute angle, find the value of $A$.

Sol.

Given, $\sec 4A = cosec(A – 20^{\circ})$

$\Rightarrow cosec(90^{\circ} – 4A) = cosec(A – 20^{\circ})$ $\quad [\because \sec \theta = cosec(90^{\circ} – \theta)]$

$\Rightarrow 90^{\circ} – 4A = A – 20^{\circ}$ $\quad [\because (90^{\circ} – 4A) \text{ and } (A – 20^{\circ}) \text{ are both acute angles}]$

$\Rightarrow 90^{\circ} – 4A = A – 20^{\circ}$

$\Rightarrow 4A + A = 90^{\circ} + 20^{\circ}$

$\Rightarrow 5A = 110^{\circ}$

$\Rightarrow A = \frac{110^{\circ}}{5}$

$\Rightarrow A = 22^{\circ}$

Therefore, the required value of $\angle A = 22^{\circ}$

6. (i) If $A, B$ and $C$ are interior angles of a triangle $ABC$, then show that $\sin\left(\frac{B+C}{2}\right) = \cos\frac{A}{2}$.

Sol.

In $\Delta ABC$, we have

$A + B + C = 180^{\circ}$ $\quad \text{[Angle sum property of triangles]}$

$B + C = 180^\circ – A$

$\Rightarrow \frac{B+C}{2} = 90^{\circ} – \frac{A}{2}$ $\quad \text{[Dividing both sides by 2]}$

$\Rightarrow \sin\left(\frac{B+C}{2}\right) = \sin\left(90^{\circ} – \frac{A}{2}\right)$ $\quad \text{[Taking sin on both sides]}$

$\Rightarrow \sin\left(\frac{B+C}{2}\right) = \cos\frac{A}{2}$ $\quad [\text{ }\because \sin(90^{\circ} – \theta) = \cos \theta]$

Hence proved

(ii) If $A, B, C$ are interior angles of $\Delta ABC$, prove that $\tan\frac{B+C}{2} = \cot\frac{A}{2}$.

Sol. In $\Delta ABC$, we have

$A + B + C = 180^{\circ}$ $\quad \text{[Angle sum property of triangles]}$

$\Rightarrow \frac{A + B + C}{2} = \frac{180^{\circ}}{2}$ $\quad \text{[Dividing both sides by 2]}$

$\Rightarrow \frac{A}{2} + \frac{B+C}{2} = 90^{\circ}$

$\Rightarrow \frac{B+C}{2} = 90^{\circ} – \frac{A}{2}$

$\Rightarrow \tan\left(\frac{B+C}{2}\right) = \tan\left(90^{\circ} – \frac{A}{2}\right)$ $\quad \text{[Taking tan on both sides]}$

$\Rightarrow \tan\left(\frac{B+C}{2}\right) = \cot\frac{A}{2}$ $\quad [\because \tan(90^{\circ} – \theta) = \cot \theta]$

Hence proved

7. Express $\sin 67^{\circ} + \cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$.

Sol. $\sin 67^{\circ} + \cos 75^{\circ}$

$= \sin(90^{\circ} – 23^{\circ}) + \cos(90^{\circ} – 15^{\circ})$

$= \cos 23^{\circ} + \sin 15^{\circ}$ $\quad [\because \sin(90^{\circ} – A) = \cos A \text{ and } \cos(90^{\circ} – A) = \sin A]$

8. (i) If $\sec 5\theta = \csc(\theta – 36^{\circ})$, where $\theta$ is an acute angle, then find the value of $\theta$.

Sol. $\sec 5\theta = \csc(\theta – 36^{\circ})$

$\Rightarrow \csc(90^{\circ} – 5\theta) = \csc(\theta – 36^{\circ})$

$\Rightarrow 90^{\circ} – 5\theta = \theta – 36^{\circ}$

$\Rightarrow 90^{\circ} + 36^{\circ} = \theta + 5\theta$

$\Rightarrow 126^{\circ} = 6\theta$

$\Rightarrow \frac{126^{\circ}}{6} = \theta$

$\Rightarrow 21^{\circ} = \theta$

$\theta = 21^{\circ}$

(ii) If $\sin A = \cos 33^{\circ}$, $A < 90$, then find the value of $A$.

Sol. $\sin A = \cos 33^{\circ}$

$\Rightarrow \cos(90^{\circ} – A) = \cos 33^{\circ}$

$\Rightarrow 90^{\circ} – A = 33^{\circ}$

$\Rightarrow 90^{\circ} – 33^{\circ} = A$

$\Rightarrow 57^{\circ} = A$

$A = 57^{\circ}$

(iii) If $\sin 2A = \cos(A + 15^{\circ})$ where $2A < 90^{\circ}$. Find the value of $A$.

Sol. $\sin 2A = \cos(A + 15^{\circ})$

$\Rightarrow \cos(90^{\circ} – 2A) = \cos(A + 15^{\circ})$

$\Rightarrow 90^{\circ} – 2A = A + 15^{\circ}$

$\Rightarrow 90^{\circ} – 15^{\circ} = A + 2A$

$\Rightarrow 75^{\circ} = 3A$

$\Rightarrow \frac{75^{\circ}}{3} = A$

$\Rightarrow 25^{\circ} = A$

$\therefore A = 25^{\circ}$

(iv) If $\sin(3x + 10) = \cos(x + 24)$, then find the value of $x$.

Sol. $\sin(3x + 10) = \cos(x + 24)$

$\Rightarrow \cos(90 – (3x + 10)) = \cos(x + 24)$

$\Rightarrow 90 – 3x – 10 = x + 24$

$\Rightarrow 80 – 3x = x + 24$

$\Rightarrow 80 – 24 = x + 3x$

$\Rightarrow 56 = 4x$

$\Rightarrow 14 = x$

$x = 14$