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

Every step explained, with hints and the method used.

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.)

Quick Solution

Q1Part (i)

EasyUnder Review

$\sin 30^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Use the 30°–60°–90° triangle.

In a 30°–60°–90° triangle with sides $1, \sqrt{3}, 2$ , the side opposite $30^{\circ}$ is 1, the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is 2. So $\sin 30^{\circ} = \frac{opposite}{hypotenuse}$ .

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

Answer

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

Quick Solution

Q1Part (ii)

EasyUnder Review

$\cos 30^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Use the 30°–60°–90° triangle.

In a 30°–60°–90° triangle with sides $1, \sqrt{3}, 2$ , the side opposite $30^{\circ}$ is 1, the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is 2. So $\cos 30^{\circ} = \frac{adjacent}{hypotenuse}$ .

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

Answer

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

Quick Solution

Q1Part (iii)

EasyUnder Review

$\tan \frac{π}{6}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Convert the angle to degrees.

Since $π rad = 180^{\circ}$ :
$\begin{matrix} \frac{π}{6} & = \frac{180^{\circ}}{6} \\ = 30^{\circ} \end{matrix}$
Use the 30°–60°–90° triangle.

In a 30°–60°–90° triangle with sides $1, \sqrt{3}, 2$ , the side opposite $30^{\circ}$ is 1, the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is 2. So $\tan 30^{\circ} = \frac{opposite}{adjacent}$ .
$\begin{matrix} \tan \frac{π}{6} & = \tan 30^{\circ} \\ = \frac{1}{\sqrt{3}} \end{matrix}$

Answer

$\tan \frac{π}{6} = \frac{1}{\sqrt{3}}$

Quick Solution

Q1Part (iv)

EasyUnder Review

$\tan 60^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Use the 30°–60°–90° triangle.

In a 30°–60°–90° triangle with sides $1, \sqrt{3}, 2$ , the side opposite $30^{\circ}$ is 1, the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is 2. So $\tan 60^{\circ} = \frac{opposite}{adjacent}$ .

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

Answer

$\tan 60^{\circ} = \sqrt{3}$

Quick Solution

Q1Part (v)

EasyUnder Review

$\sec 60^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Write sec as a reciprocal.

By definition $\sec θ = \frac{1}{\cos θ}$ , and $\cos 60^{\circ} = \frac{1}{2}$ .

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

Answer

$\sec 60^{\circ} = 2$

Quick Solution

Q1Part (vi)

EasyUnder Review

$\cos \frac{π}{3}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Convert the angle to degrees.

Since $π rad = 180^{\circ}$ :
$\begin{matrix} \frac{π}{3} & = \frac{180^{\circ}}{3} \\ = 60^{\circ} \end{matrix}$
Use the 30°–60°–90° triangle.

The side adjacent to $60^{\circ}$ is 1 and the hypotenuse is 2.
$\begin{matrix} \cos \frac{π}{3} & = \cos 60^{\circ} \\ = \frac{1}{2} \end{matrix}$

Answer

$\cos \frac{π}{3} = \frac{1}{2}$

Quick Solution

Q1Part (vii)

EasyUnder Review

$\cot 60^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Write cot as a reciprocal.

By definition $\cot θ = \frac{1}{\tan θ}$ , and $\tan 60^{\circ} = \sqrt{3}$ .

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

Answer

$\cot 60^{\circ} = \frac{1}{\sqrt{3}}$

Quick Solution

Q1Part (viii)

EasyUnder Review

$\sin 60^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Use the 30°–60°–90° triangle.

In a 30°–60°–90° triangle with sides $1, \sqrt{3}, 2$ , the side opposite $30^{\circ}$ is 1, the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is 2. So $\sin 60^{\circ} = \frac{opposite}{hypotenuse}$ .

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

Answer

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

Quick Solution

Q1Part (ix)

EasyUnder Review

$\sec 30^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Write sec as a reciprocal.

By definition $\sec θ = \frac{1}{\cos θ}$ , and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ .

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

Answer

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

Quick Solution

Q1Part (x)

EasyUnder Review

$cosec 30^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Write cosec as a reciprocal.

By definition $cosec θ = \frac{1}{\sin θ}$ , and $\sin 30^{\circ} = \frac{1}{2}$ .

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

Answer

$cosec 30^{\circ} = 2$

Quick Solution

Q1Part (xi)

EasyUnder Review

$\sin 45^{\circ}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Use the 45°–45°–90° triangle.

In a 45°–45°–90° triangle with legs 1 and 1, the hypotenuse is $\sqrt{2}$ . So $\sin 45^{\circ} = \frac{opposite}{hypotenuse}$ .

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

Answer

$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$

Quick Solution

Q1Part (xii)

EasyUnder Review

$\cos \frac{π}{4}$

Hint

Use the exact special-angle value; for a reciprocal ratio, take 1 over the basic ratio.

Solution

Convert the angle to degrees.

Since $π rad = 180^{\circ}$ :
$\begin{matrix} \frac{π}{4} & = \frac{180^{\circ}}{4} \\ = 45^{\circ} \end{matrix}$
Use the 45°–45°–90° triangle.

In a 45°–45°–90° triangle with legs 1 and 1, the hypotenuse is $\sqrt{2}$ . So $\cos 45^{\circ} = \frac{adjacent}{hypotenuse}$ .
$\begin{matrix} \cos \frac{π}{4} & = \cos 45^{\circ} \\ = \frac{1}{\sqrt{2}} \end{matrix}$

Answer

$\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$

Quick Solution

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.)

Quick Solution

Q2Part (i)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$ .
$\begin{matrix} 2 \sin 60^{\circ} \cos 60^{\circ} & = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \end{matrix}$
Simplify.

Cancel the 2 in the numerator with one 2 in the denominator.
$\begin{matrix} 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} & = \frac{2 \sqrt{3}}{4} \\ = \frac{\sqrt{3}}{2} \end{matrix}$

Answer

$2 \sin 60^{\circ} \cos 60^{\circ} = \frac{\sqrt{3}}{2}$

Quick Solution

Q2Part (ii)

EasyUnder Review

$2 \cos \frac{π}{6} \sin \frac{π}{6}$

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Convert the angle to degrees.

Since $π rad = 180^{\circ}$ :
$\begin{matrix} \frac{π}{6} & = 30^{\circ} \end{matrix}$
Substitute the exact values.

Use $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} 2 \cos \frac{π}{6} \sin \frac{π}{6} & = 2 \cos 30^{\circ} \sin 30^{\circ} \\ = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \end{matrix}$
Simplify.

Cancel the common factor 2.
$\begin{matrix} 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} & = \frac{\sqrt{3}}{2} \end{matrix}$

Answer

$2 \cos \frac{π}{6} \sin \frac{π}{6} = \frac{\sqrt{3}}{2}$

Quick Solution

Q2Part (iii)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ .
$\begin{matrix} 2 \sin 45^{\circ} + 2 \cos 45^{\circ} & = 2 \times \frac{1}{\sqrt{2}} + 2 \times \frac{1}{\sqrt{2}} \\ = \frac{4}{\sqrt{2}} \end{matrix}$
Simplify the surd.

Write $4 = 2 \times \sqrt{2} \times \sqrt{2}$ and cancel one $\sqrt{2}$ .
$\begin{matrix} \frac{4}{\sqrt{2}} & = \frac{2 \sqrt{2} \sqrt{2}}{\sqrt{2}} \\ = 2 \sqrt{2} \end{matrix}$

Answer

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

Quick Solution

Q2Part (iv)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} \sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ} & = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} \end{matrix}$
Multiply and add.

Use $\sqrt{3} \cdot \sqrt{3} = 3$ , then add the like fractions.
$\begin{matrix} \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} & = \frac{3}{4} + \frac{1}{4} \\ = \frac{4}{4} \\ = 1 \end{matrix}$

Answer

$\sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ} = 1$

Quick Solution

Q2Part (v)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} \cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ} & = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \end{matrix}$
Multiply and subtract.

Both products equal $\frac{\sqrt{3}}{4}$ , so they cancel.
$\begin{matrix} \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2} & = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\ = 0 \end{matrix}$

Answer

$\cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ} = 0$

Quick Solution

Q2Part (vi)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} \sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ} & = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \frac{1}{2} \end{matrix}$
Multiply and subtract.

Use $\sqrt{3} \cdot \sqrt{3} = 3$ , then subtract the like fractions.
$\begin{matrix} \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \frac{1}{2} & = \frac{3}{4} - \frac{1}{4} \\ = \frac{2}{4} \\ = \frac{1}{2} \end{matrix}$

Answer

$\sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ} = \frac{1}{2}$

Quick Solution

Q2Part (vii)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Substitute the exact values.

Use $\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} \cos 60^{\circ} \cos 30^{\circ} + \sin 60^{\circ} \sin 30^{\circ} & = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \end{matrix}$
Multiply and add.

Both products equal $\frac{\sqrt{3}}{4}$ .
$\begin{matrix} \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} & = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ = \frac{2 \sqrt{3}}{4} \\ = \frac{\sqrt{3}}{2} \end{matrix}$

Answer

$\cos 60^{\circ} \cos 30^{\circ} + \sin 60^{\circ} \sin 30^{\circ} = \frac{\sqrt{3}}{2}$

Quick Solution

Q2Part (viii)

EasyUnder Review

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

Hint

Substitute the exact values of the special-angle ratios, then simplify.

Solution

Convert the angle to degrees.

Since $π rad = 180^{\circ}$ :
$\begin{matrix} \frac{π}{6} & = 30^{\circ} \end{matrix}$
Substitute the exact values.

Use $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\cot 30^{\circ} = \frac{1}{\tan 30^{\circ}} = \sqrt{3}$ .
$\begin{matrix} \tan \frac{π}{6} \cot \frac{π}{6} + 1 & = \tan 30^{\circ} \cot 30^{\circ} + 1 \\ = \frac{1}{\sqrt{3}} \cdot \sqrt{3} + 1 \end{matrix}$
Simplify.

The product of a ratio and its reciprocal is 1.
$\begin{matrix} \frac{1}{\sqrt{3}} \cdot \sqrt{3} + 1 & = 1 + 1 \\ = 2 \end{matrix}$

Answer

$\tan \frac{π}{6} \cot \frac{π}{6} + 1 = 2$

Quick Solution

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.)

Quick Solution

Q3Part (i)

EasyUnder Review

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

Hint

Substitute $\frac{1}{\sqrt{2}}$ for the 45° ratios and simplify.

Solution

Substitute the exact values.

Use $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ .

\begin{matrix} 2 \sin 45^{\circ} - 2 \cos 45^{\circ} & = 2 \times \frac{1}{\sqrt{2}} - 2 \times \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} \\ = 0 \end{matrix}

Answer

$2 \sin 45^{\circ} - 2 \cos 45^{\circ} = 0$

Quick Solution

Q3Part (ii)

EasyUnder Review

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

Hint

Substitute $\frac{1}{\sqrt{2}}$ for the 45° ratios and simplify.

Solution

Substitute the exact values.

Use $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and, from the 30°–60°–90° triangle, $\sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} 3 \cos 45^{\circ} + 4 \sin 30^{\circ} & = 3 \times \frac{1}{\sqrt{2}} + 4 \times \frac{1}{2} \\ = \frac{3}{\sqrt{2}} + 2 \end{matrix}$
Rationalise the denominator.

Multiply the first term by $\frac{\sqrt{2}}{\sqrt{2}}$ , then write over a common denominator.
$\begin{matrix} \frac{3}{\sqrt{2}} + 2 & = \frac{3 \sqrt{2}}{2} + 2 \\ = \frac{3 \sqrt{2} + 4}{2} \end{matrix}$

Answer

$3 \cos 45^{\circ} + 4 \sin 30^{\circ} = \frac{3 \sqrt{2} + 4}{2}$

Quick Solution

Q3Part (iii)

EasyUnder Review

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

Hint

Substitute $\frac{1}{\sqrt{2}}$ for the 45° ratios and simplify.

Solution

Substitute the exact values.

Use $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ .
$\begin{matrix} 5 \cos 45^{\circ} - 3 \sin 45^{\circ} & = \frac{5}{\sqrt{2}} - \frac{3}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} \end{matrix}$
Simplify the surd.

Since $2 = \sqrt{2} \times \sqrt{2}$ , one $\sqrt{2}$ cancels.
$\begin{matrix} \frac{2}{\sqrt{2}} & = \frac{\sqrt{2} \sqrt{2}}{\sqrt{2}} \\ = \sqrt{2} \end{matrix}$

Answer

$5 \cos 45^{\circ} - 3 \sin 45^{\circ} = \sqrt{2}$

Quick Solution

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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