Mathematics

Class 92025–26 SNC17 Questions SolvedEnglish Medium

Exercise 6.3 — Class 9 Mathematics Solutions

All 17 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.3

12 Questions (5 Sub-parts)

Quick Solution Step by Step

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

Question 1

EasyUnder Review

If $θ$ lies in the first quadrant, find the remaining trigonometric ratios of $θ$ .

Hint

Turn the given ratio into two sides of a right-angled triangle, find the third side by Pythagoras theorem, then read off every ratio (all positive in the first quadrant).

Solution

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

Step by Step

Q1Part (i)

Easy

$\sin θ = \frac{2}{3}$

Quick Solution

Under Review

\begin{matrix} B^{2} & = H^{2} - P^{2} \\ = 3^{2} - 2^{2} \\ = 9 - 4 \\ = 5 \\ ⟹ B & = \sqrt{5} \\ \cos θ & = \frac{B}{H} \\ = \frac{\sqrt{5}}{3} \\ \tan θ & = \frac{P}{B} \\ = \frac{2}{\sqrt{5}} \\ = \frac{2 \sqrt{5}}{5} \\ cosec θ & = \frac{H}{P} \\ = \frac{3}{2} \\ \sec θ & = \frac{H}{B} \\ = \frac{3}{\sqrt{5}} \\ = \frac{3 \sqrt{5}}{5} \\ \cot θ & = \frac{B}{P} \\ = \frac{\sqrt{5}}{2} \end{matrix}

Step by Step

Q1Part (ii)

Easy

$\cos θ = \frac{3}{4}$

Quick Solution

Under Review

\begin{matrix} P^{2} & = H^{2} - B^{2} \\ = 4^{2} - 3^{2} \\ = 16 - 9 \\ = 7 \\ ⟹ P & = \sqrt{7} \\ \sin θ & = \frac{P}{H} \\ = \frac{\sqrt{7}}{4} \\ \tan θ & = \frac{P}{B} \\ = \frac{\sqrt{7}}{3} \\ cosec θ & = \frac{H}{P} \\ = \frac{4}{\sqrt{7}} \\ = \frac{4 \sqrt{7}}{7} \\ \sec θ & = \frac{H}{B} \\ = \frac{4}{3} \\ \cot θ & = \frac{B}{P} \\ = \frac{3}{\sqrt{7}} \\ = \frac{3 \sqrt{7}}{7} \end{matrix}

Step by Step

Q1Part (iii)

Easy

$\tan θ = \frac{1}{2}$

Quick Solution

Under Review

\begin{matrix} H^{2} & = P^{2} + B^{2} \\ = 1^{2} + 2^{2} \\ = 1 + 4 \\ = 5 \\ ⟹ H & = \sqrt{5} \\ \sin θ & = \frac{P}{H} \\ = \frac{1}{\sqrt{5}} \\ = \frac{\sqrt{5}}{5} \\ \cos θ & = \frac{B}{H} \\ = \frac{2}{\sqrt{5}} \\ = \frac{2 \sqrt{5}}{5} \\ cosec θ & = \frac{H}{P} \\ = \sqrt{5} \\ \sec θ & = \frac{H}{B} \\ = \frac{\sqrt{5}}{2} \\ \cot θ & = \frac{B}{P} \\ = 2 \end{matrix}

Step by Step

Q1Part (iv)

Easy

$\sec θ = 3$

Quick Solution

Under Review

\begin{matrix} P^{2} & = H^{2} - B^{2} \\ = 3^{2} - 1^{2} \\ = 9 - 1 \\ = 8 \\ ⟹ P & = \sqrt{8} \\ = 2 \sqrt{2} \\ \sin θ & = \frac{P}{H} \\ = \frac{2 \sqrt{2}}{3} \\ \cos θ & = \frac{B}{H} \\ = \frac{1}{3} \\ \tan θ & = \frac{P}{B} \\ = 2 \sqrt{2} \\ cosec θ & = \frac{H}{P} \\ = \frac{3}{2 \sqrt{2}} \\ = \frac{3 \sqrt{2}}{4} \\ \cot θ & = \frac{B}{P} \\ = \frac{1}{2 \sqrt{2}} \\ = \frac{\sqrt{2}}{4} \end{matrix}

Step by Step

Q1Part (v)

Medium

$\cot θ = \sqrt{\frac{3}{2}}$

Quick Solution

Under Review

\begin{matrix} H^{2} & = P^{2} + B^{2} \\ = (\sqrt{2})^{2} + (\sqrt{3})^{2} \\ = 2 + 3 \\ = 5 \\ ⟹ H & = \sqrt{5} \\ \sin θ & = \frac{P}{H} \\ = \frac{\sqrt{2}}{\sqrt{5}} \\ = \frac{\sqrt{10}}{5} \\ \cos θ & = \frac{B}{H} \\ = \frac{\sqrt{3}}{\sqrt{5}} \\ = \frac{\sqrt{15}}{5} \\ \tan θ & = \frac{P}{B} \\ = \frac{\sqrt{2}}{\sqrt{3}} \\ = \frac{\sqrt{6}}{3} \\ cosec θ & = \frac{H}{P} \\ = \frac{\sqrt{5}}{\sqrt{2}} \\ = \frac{\sqrt{10}}{2} \\ \sec θ & = \frac{H}{B} \\ = \frac{\sqrt{5}}{\sqrt{3}} \\ = \frac{\sqrt{15}}{3} \end{matrix}

Step by Step

Question 2

Easy

Prove that: $(\sin θ + \cos θ)^{2} = 1 + 2 \sin θ \cos θ$

Quick Solution

Under Review

\begin{matrix} L.H.S & = (\sin θ + \cos θ)^{2} \\ = \sin^{2} θ + 2 \sin θ \cos θ + \cos^{2} θ \\ = (\sin^{2} θ + \cos^{2} θ) + 2 \sin θ \cos θ \\ = 1 + 2 \sin θ \cos θ \\ = R.H.S \end{matrix}

Step by Step

Question 3

Easy

Prove that: $\frac{\cos θ}{\sin θ} = \frac{1}{\tan θ}$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \frac{\cos θ}{\sin θ} \\ = \cot θ \\ = \frac{1}{\tan θ} \\ = R.H.S \end{matrix}

Step by Step

Question 4

Easy

Prove that: $\frac{\sin θ}{cosec θ} + \frac{\cos θ}{\sec θ} = 1$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \frac{\sin θ}{cosec θ} + \frac{\cos θ}{\sec θ} \\ = \sin θ \cdot \sin θ + \cos θ \cdot \cos θ \\ = \sin^{2} θ + \cos^{2} θ \\ = 1 \\ = R.H.S \end{matrix}

Step by Step

Question 5

Easy

Prove that: $\cos^{2} θ - \sin^{2} θ = 2 \cos^{2} θ - 1$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \cos^{2} θ - \sin^{2} θ \\ = \cos^{2} θ - (1 - \cos^{2} θ) \\ = \cos^{2} θ - 1 + \cos^{2} θ \\ = 2 \cos^{2} θ - 1 \\ = R.H.S \end{matrix}

Step by Step

Question 6

Easy

Prove that: $\cos^{2} θ - \sin^{2} θ = 1 - 2 \sin^{2} θ$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \cos^{2} θ - \sin^{2} θ \\ = (1 - \sin^{2} θ) - \sin^{2} θ \\ = 1 - 2 \sin^{2} θ \\ = R.H.S \end{matrix}

Step by Step

Question 7

Medium

Prove that: $\frac{1 - \sin θ}{\cos θ} = \frac{\cos θ}{1 + \sin θ}$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \frac{1 - \sin θ}{\cos θ} \times \frac{1 + \sin θ}{1 + \sin θ} \\ = \frac{1 - \sin^{2} θ}{\cos θ (1 + \sin θ)} \\ = \frac{\cos^{2} θ}{\cos θ (1 + \sin θ)} \\ = \frac{\cos θ}{1 + \sin θ} \\ = R.H.S \end{matrix}

Step by Step

Question 8

Medium

Prove that: $(\sec θ - \tan θ)^{2} = \frac{1 - \sin θ}{1 + \sin θ}$

Quick Solution

Under Review

\begin{matrix} L.H.S & = (\sec θ - \tan θ)^{2} \\ = {(\frac{1}{\cos θ} - \frac{\sin θ}{\cos θ})}^{2} \\ = \frac{(1 - \sin θ)^{2}}{\cos^{2} θ} \\ = \frac{(1 - \sin θ)^{2}}{1 - \sin^{2} θ} \\ = \frac{(1 - \sin θ)^{2}}{(1 - \sin θ) (1 + \sin θ)} \\ = \frac{1 - \sin θ}{1 + \sin θ} \\ = R.H.S \end{matrix}

Step by Step

Question 9

Medium

Prove that: $(\tan θ + \cot θ)^{2} = \sec^{2} θ {cosec}^{2} θ$

Quick Solution

Under Review

\begin{matrix} L.H.S & = (\tan θ + \cot θ)^{2} \\ = {(\frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ})}^{2} \\ = {(\frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ})}^{2} \\ = {(\frac{1}{\sin θ \cos θ})}^{2} \\ = \frac{1}{\cos^{2} θ} \cdot \frac{1}{\sin^{2} θ} \\ = \sec^{2} θ {cosec}^{2} θ \\ = R.H.S \end{matrix}

Step by Step

Question 10

Hard

Prove that: $\frac{\tan θ + \sec θ - 1}{\tan θ - \sec θ + 1} = \tan θ + \sec θ$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \frac{\tan θ + \sec θ - 1}{\tan θ - \sec θ + 1} \\ = \frac{(\tan θ + \sec θ) - (\sec^{2} θ - \tan^{2} θ)}{\tan θ - \sec θ + 1} \\ = \frac{(\tan θ + \sec θ) - (\sec θ + \tan θ) (\sec θ - \tan θ)}{\tan θ - \sec θ + 1} \\ = \frac{(\tan θ + \sec θ) [1 - \sec θ + \tan θ]}{\tan θ - \sec θ + 1} \\ = \frac{(\tan θ + \sec θ) (\tan θ - \sec θ + 1)}{\tan θ - \sec θ + 1} \\ = \tan θ + \sec θ \\ = R.H.S \end{matrix}

Step by Step

Question 11

Medium

Prove that: $\sin^{3} θ - \cos^{3} θ = (\sin θ - \cos θ) (1 + \sin θ \cos θ)$

Quick Solution

Under Review

\begin{matrix} L.H.S & = \sin^{3} θ - \cos^{3} θ \\ = (\sin θ - \cos θ) (\sin^{2} θ + \sin θ \cos θ + \cos^{2} θ) \\ = (\sin θ - \cos θ) (1 + \sin θ \cos θ) \\ = R.H.S \end{matrix}

Step by Step

Question 12

Hard

Prove that: $\sin^{6} θ - \cos^{6} θ = (\sin^{2} θ - \cos^{2} θ) (1 - \sin^{2} θ \cos^{2} θ)$

Quick Solution

Under Review

\begin{matrix} L.H.S & = (\sin^{2} θ)^{3} - (\cos^{2} θ)^{3} \\ = (\sin^{2} θ - \cos^{2} θ) [\sin^{4} θ + \sin^{2} θ \cos^{2} θ + \cos^{4} θ] \\ = (\sin^{2} θ - \cos^{2} θ) [(\sin^{2} θ + \cos^{2} θ)^{2} - \sin^{2} θ \cos^{2} θ] \\ = (\sin^{2} θ - \cos^{2} θ) [1 - \sin^{2} θ \cos^{2} θ] \\ = R.H.S \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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