Mathematics

Class 92025–26 SNC26 Questions SolvedEnglish Medium

Exercise 6.4 — Class 9 Mathematics Solutions

All 26 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.4

3 Questions (23 Sub-parts)

Quick Solution Step by Step

Just the working. Switch to Step by Step for the full explanation.

Question 1

EasyUnder Review

Find the value of the following trigonometric ratios without using the calculator.

Hint

Convert any radian angle to degrees, then use the exact values for $30^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ (taking reciprocals for $\sec$ , $cosec$ and $\cot$ ).

Solution

(This question consists of multiple subparts. See solutions below.)

Step by Step

Q1Part (i)

Easy

$\sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} \sin 30^{\circ} & = \frac{1}{2} \end{matrix}

Step by Step

Q1Part (ii)

Easy

$\cos 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} \cos 30^{\circ} & = \frac{\sqrt{3}}{2} \end{matrix}

Step by Step

Q1Part (iii)

Easy

$\tan \frac{π}{6}$

Quick Solution

Under Review

\begin{matrix} \tan \frac{π}{6} & = \tan 30^{\circ} \\ = \frac{1}{\sqrt{3}} \end{matrix}

Step by Step

Q1Part (iv)

Easy

$\tan 60^{\circ}$

Quick Solution

Under Review

\begin{matrix} \tan 60^{\circ} & = \frac{\sqrt{3}}{1} \\ = \sqrt{3} \end{matrix}

Step by Step

Q1Part (v)

Easy

$\sec 60^{\circ}$

Quick Solution

Under Review

\begin{matrix} \sec 60^{\circ} & = \frac{1}{\cos 60^{\circ}} \\ = \frac{1}{1 / 2} \\ = 2 \end{matrix}

Step by Step

Q1Part (vi)

Easy

$\cos \frac{π}{3}$

Quick Solution

Under Review

\begin{matrix} \cos \frac{π}{3} & = \cos 60^{\circ} \\ = \frac{1}{2} \end{matrix}

Step by Step

Q1Part (vii)

Easy

$\cot 60^{\circ}$

Quick Solution

Under Review

\begin{matrix} \cot 60^{\circ} & = \frac{1}{\tan 60^{\circ}} \\ = \frac{1}{\sqrt{3}} \end{matrix}

Step by Step

Q1Part (viii)

Easy

$\sin 60^{\circ}$

Quick Solution

Under Review

\begin{matrix} \sin 60^{\circ} & = \frac{\sqrt{3}}{2} \end{matrix}

Step by Step

Q1Part (ix)

Easy

$\sec 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} \sec 30^{\circ} & = \frac{1}{\cos 30^{\circ}} \\ = \frac{1}{\sqrt{3} / 2} \\ = \frac{2}{\sqrt{3}} \end{matrix}

Step by Step

Q1Part (x)

Easy

$cosec 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} cosec 30^{\circ} & = \frac{1}{\sin 30^{\circ}} \\ = \frac{1}{1 / 2} \\ = 2 \end{matrix}

Step by Step

Q1Part (xi)

Easy

$\sin 45^{\circ}$

Quick Solution

Under Review

\begin{matrix} \sin 45^{\circ} & = \frac{1}{\sqrt{2}} \end{matrix}

Step by Step

Q1Part (xii)

Easy

$\cos \frac{π}{4}$

Quick Solution

Under Review

\begin{matrix} \cos \frac{π}{4} & = \cos 45^{\circ} \\ = \frac{1}{\sqrt{2}} \end{matrix}

Step by Step

Question 2

EasyUnder Review

Evaluate.

Hint

Replace each ratio by its exact special-angle value, then simplify the fractions and surds.

Solution

(This question consists of multiple subparts. See solutions below.)

Step by Step

Q2Part (i)

Easy

$2 \sin 60^{\circ} \cos 60^{\circ}$

Quick Solution

Under Review

\begin{matrix} 2 \sin 60^{\circ} \cos 60^{\circ} & = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ = \frac{\sqrt{3}}{2} \end{matrix}

Step by Step

Q2Part (ii)

Easy

$2 \cos \frac{π}{6} \sin \frac{π}{6}$

Quick Solution

Under Review

\begin{matrix} 2 \cos \frac{π}{6} \sin \frac{π}{6} & = 2 \cos 30^{\circ} \sin 30^{\circ} \\ = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ = \frac{\sqrt{3}}{2} \end{matrix}

Step by Step

Q2Part (iii)

Easy

$2 \sin 45^{\circ} + 2 \cos 45^{\circ}$

Quick Solution

Under Review

\begin{matrix} 2 \sin 45^{\circ} + 2 \cos 45^{\circ} & = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} \\ = \frac{4}{\sqrt{2}} \\ = 2 \sqrt{2} \end{matrix}

Step by Step

Q2Part (iv)

Easy

$\sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} Expr & = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} \\ = \frac{3}{4} + \frac{1}{4} \\ = 1 \end{matrix}

Step by Step

Q2Part (v)

Easy

$\cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} Expr & = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\ = 0 \end{matrix}

Step by Step

Q2Part (vi)

Easy

$\sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} Expr & = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \frac{1}{2} \\ = \frac{3}{4} - \frac{1}{4} \\ = \frac{1}{2} \end{matrix}

Step by Step

Q2Part (vii)

Easy

$\cos 60^{\circ} \cos 30^{\circ} + \sin 60^{\circ} \sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} Expr & = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ = \frac{2 \sqrt{3}}{4} \\ = \frac{\sqrt{3}}{2} \end{matrix}

Step by Step

Q2Part (viii)

Easy

$\tan \frac{π}{6} \cot \frac{π}{6} + 1$

Quick Solution

Under Review

\begin{matrix} Expr & = \tan 30^{\circ} \cot 30^{\circ} + 1 \\ = \frac{1}{\sqrt{3}} \cdot \sqrt{3} + 1 \\ = 2 \end{matrix}

Step by Step

Question 3

EasyUnder Review

If $\sin 45^{\circ}$ and $\cos 45^{\circ}$ are each equal to $\frac{1}{\sqrt{2}}$ , then find the value of the following:

Hint

Put $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ (and any other special-angle values needed), then simplify the surds.

Solution

(This question consists of multiple subparts. See solutions below.)

Step by Step

Q3Part (i)

Easy

$2 \sin 45^{\circ} - 2 \cos 45^{\circ}$

Quick Solution

Under Review

\begin{matrix} 2 \sin 45^{\circ} - 2 \cos 45^{\circ} & = \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} \\ = 0 \end{matrix}

Step by Step

Q3Part (ii)

Easy

$3 \cos 45^{\circ} + 4 \sin 30^{\circ}$

Quick Solution

Under Review

\begin{matrix} 3 \cos 45^{\circ} + 4 \sin 30^{\circ} & = 3 \cdot \frac{1}{\sqrt{2}} + 4 \cdot \frac{1}{2} \\ = \frac{3 \sqrt{2}}{2} + 2 \\ = \frac{3 \sqrt{2} + 4}{2} \end{matrix}

Step by Step

Q3Part (iii)

Easy

$5 \cos 45^{\circ} - 3 \sin 45^{\circ}$

Quick Solution

Under Review

\begin{matrix} 5 \cos 45^{\circ} - 3 \sin 45^{\circ} & = \frac{5}{\sqrt{2}} - \frac{3}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} \\ = \sqrt{2} \end{matrix}

Step by Step

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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