Q2: In Fig. 8.13, find $\tan P – \cot R$.

Solution:

In right-angled $\Delta PQR$,

$ PQ = 12 cm $

$ PR = 13 cm $

By using the Pythagoras Theorem, we have:

$PR^2 = PQ^2 + QR^2$

$\implies (13)^2 = (12)^2 + QR^2$

$\implies 169 = 144 + QR^2$

$\implies QR^2 = 169 – 144$

$\implies QR^2 = 25$

$\implies QR = \sqrt{25}$

$\therefore QR = 5\text{ cm}$

Now,

$\tan P =  \frac{QR}{PQ} = \frac{5}{12}$

$\cot R =  \frac{QR}{PQ} = \frac{5}{12}$

Therefore,

$\tan P – \cot R = \frac{5}{12} – \frac{5}{12} = 0$

Q3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.

Solution:

Let  $ABC$ be a right-angled triangle, where $\angle B = 90^\circ$

We know that:

$\sin A = \frac{BC}{AC} = \frac{3}{4}$

Therefore, if $BC = 3k$, then $AC = 4k$, where $k$ is a positive number.

Now, by using the Pythagoras Theorem, we have:

$AB^2 = AC^2 – BC^2$

$\implies AB^2 = (4k)^2 – (3k)^2$

$\implies AB^2 = 16k^2 – 9k^2$

$\implies AB^2 = 7k^2$

$\implies AB = \sqrt{7k^2}$

$\therefore AB = \sqrt{7}k$

Therefore,

$\cos A = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$

$\tan A = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$

Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.

Solution:

Let $ABC$ be a right-angled triangle, where $\angle B = 90^\circ$.

Given that $15 \cot A = 8$

$\implies \cot A = \frac{8}{15}$

We know that:

$\cot A = \frac{AB}{BC} = \frac{8}{15}$

Therefore, if $AB = 8k$, then $BC = 15k$, where $k$ is a positive number.

By using the Pythagoras Theorem, we have:

$AC^2 = AB^2 + BC^2$

$\implies AC^2 = (8k)^2 + (15k)^2$

$\implies AC^2 = 64k^2 + 225k^2$

$\implies AC^2 = 289k^2$

$\implies AC = \sqrt{289k^2}$

$\therefore AC = 17k$

∴ $\sin A = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}$

$\sec A = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}$

Q5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.

Solution:

Let us consider a right-angled $\Delta ABC$ where $\angle B = 90^\circ$ and $\angle A = \theta$.

We know that:

$\sec \theta = \frac{AC}{AB} = \frac{13}{12}$

Therefore, if $AC = 13k$, then $AB = 12k$, where $k$ is a positive number.

By using the Pythagoras Theorem, we have:

$BC^2 = AC^2 – AB^2$

$\implies BC^2 = (13k)^2 – (12k)^2$

$\implies BC^2 = 169k^2 – 144k^2$

$\implies BC^2 = 25k^2$

$\implies BC = \sqrt{25k^2}$

$\therefore BC = 5k$

Now,

$\sin \theta = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$

$\cos \theta = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$

$\tan \theta = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$

$\csc \theta = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$

$\cot \theta = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$

Q6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.

Solution:

Let  $ABC$ be a right-angled triangle, where $\angle c = 90^\circ$  so that $\angle A$ and $\angle B$ are its acute angles.

Now,

$\cos A = \frac{AC}{AB}$

$\cos B = \frac{BC}{AB}$

We are given that:

$\cos A = \cos B$

$\implies \frac{AC}{AB} = \frac{BC}{AB}$

$\implies AC = BC$

Since angles opposite to equal sides in a triangle are equal:

$\therefore \angle A = \angle B$

Hence proved.

Q7: If $\cot \theta = \frac{7}{8}$, evaluate:

(i) $\frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)}$

(ii) $\cot^2 \theta$

Solution:

Let  $ABC$ be a right-angled triangle, where $\angle B = 90^\circ$ and $\angle A = \theta$.

We know that:

$\cot \theta = \frac{AB}{BC} = \frac{7}{8}$

Therefore, if $AB = 7k$, then $BC = 8k$, where $k$ is a positive number.

By using the Pythagoras Theorem, we have:

$AC^2 = AB^2 + BC^2$

$\implies AC^2 = (7k)^2 + (8k)^2$

$\implies AC^2 = 49k^2 + 64k^2$

$\implies AC^2 = 113k^2$

$\implies AC = \sqrt{113k^2}$

$\therefore AC = \sqrt{113}k$

So, we find:

$\sin \theta = \frac{BC}{AC} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$

$\cos \theta = \frac{AB}{AC} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}$

(i) $\frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)}$

$ = \frac{1 – \sin^2 \theta}{1 – \cos^2 \theta} $

$ = \frac{1 – \left(\frac{8}{\sqrt{113}}\right)^2}{1 – \left(\frac{7}{\sqrt{113}}\right)^2}$

$ = \frac{1 – \frac{64}{113}}{1 – \frac{49}{113}} $

$ = \frac{\frac{113 – 64}{113}}{\frac{113 – 49}{113}}$

$\therefore \frac{49}{64}$

(ii) $\cot^2 \theta$

$\cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2$

$\therefore \cot^2 \theta = \frac{49}{64}$

Q8: If $3 \cot A = 4$, check whether $\frac{1 – \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.

Solution:

Let $ABC$ be a right-angled triangle, where $\angle B = 90^\circ$.

We are given $3 \cot A = 4 $

$ \implies \cot A = \frac{4}{3}$.

We know that:

$\cot A = \frac{AB}{BC} = \frac{4}{3}$

Therefore, if $AB = 4k$, then $BC = 3k$, where $k$ is a positive number.

By using the Pythagoras Theorem, we have:

$AC^2 = AB^2 + BC^2$

$\implies AC^2 = (4k)^2 + (3k)^2$

$\implies AC^2 = 16k^2 + 9k^2$

$\implies AC^2 = 25k^2$

$\implies AC = \sqrt{25k^2}$

$\therefore AC = 5k$

From this, we get:

$\tan A = \frac{BC}{AB} = \frac{3}{4}$,

$\cos A = \frac{AB}{AC} = \frac{4}{5}$,

$\sin A = \frac{BC}{AC} = \frac{3}{5}$

Now, $\text{LHS}$

$= \frac{1 – \tan^2 A}{1 + \tan^2 A}$

$ = \frac{1 – \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2}$

$= \frac{1 – \frac{9}{16}}{1 + \frac{9}{16}}$

$ = \frac{\frac{16 – 9}{16}}{\frac{16 + 9}{16}}$

$\therefore \text{LHS} = \frac{7}{25}$

Now,  $\text{RHS}$

$= \cos^2 A – \sin^2 A $

$ = \left(\frac{4}{5}\right)^2 – \left(\frac{3}{5}\right)^2$

$ = \frac{16}{25} – \frac{9}{25}$

$\therefore \text{RHS} = \frac{7}{25}$

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

Q9: In triangle $ABC$, right-angled at $B$, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:

(i) $\sin A \cos C + \cos A \sin C$

(ii) $\cos A \cos C – \sin A \sin C$

Solution:

Let $ABC$ be a right-angled triangle, where $\angle B = 90^\circ$.

Given $\tan A = \frac{1}{\sqrt{3}}$

We know that, $\tan A = \frac{BC}{AB} = \frac{1}{\sqrt{3}}$.

Let $BC = k$ and $AB = \sqrt{3}k$.

By using the Pythagoras Theorem, we have:

$AC^2 = AB^2 + BC^2$

$\implies AC^2 = (\sqrt{3}k)^2 + (k)^2$

$\implies AC^2 = 3k^2 + k^2$

$\implies AC^2 = 4k^2$

$\implies AC = \sqrt{4k^2}$

$\therefore AC = 2k$

Now,

$\sin A = \frac{BC}{AC} = \frac{1}{2}$

$\cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$

$\sin C = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$

$\cos C = \frac{BC}{AC} = \frac{1}{2}$

(i) $\sin A \cos C + \cos A \sin C$

$= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

$\frac{1}{4} + \frac{3}{4} = \frac{4}{4}$

$\therefore \sin A \cos C + \cos A \sin C = 1$

(ii) $\cos A \cos C – \sin A \sin C$

$= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) – \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = 0$

∴ $\cos A \cos C – \sin A \sin C = 0 $

Q10: In $\Delta PQR$, right-angled at $Q$, $PR + QR = 25\text{ cm}$ and $PQ = 5\text{ cm}$. Determine the values of $\sin P$, $\cos P$ and $\tan P$.

Solution:

Let us consider right-angled $\Delta PQR$.

We are given $PR + QR = 25 $

$ \implies PR = 25 – QR$.

By using the Pythagoras Theorem, we have:

$PR^2 = PQ^2 + QR^2$

$\implies (25 – QR)^2 = (5)^2 + QR^2$

$\implies 625 + QR^2 – 50QR = 25 + QR^2$

$\implies 625 – 25 = 50QR$

$\implies 600 = 50QR$

$\implies QR = \frac{600}{50}$

$\therefore QR = 12\text{ cm}$

Substituting $QR$ value to find $PR$:

$PR = 25 – 12$

$\therefore PR = 13\text{ cm}$

Now,

$\sin P = \frac{QR}{PR} = \frac{12}{13}$

$\cos P = \frac{PQ}{PR} = \frac{5}{13}$

$\tan P = \frac{QR}{PQ} = \frac{12}{5}$

Q11: State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.

False. $\tan A = \frac{\text{side opposite}}{\text{side adjacent}}$. If the side opposite is longer than the side adjacent, $\tan A$ will be greater than 1. For example, $\tan 60^\circ = \sqrt{3} \approx 1.732$.

(ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.

True. As $\sec A = \frac{\text{hypotenuse}}{\text{side adjacent}}$, the value must always be greater than or equal to 1 since the hypotenuse is the longest side. Here $\frac{12}{5} = 2.4 > 1$.

(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.

False. $\cos A$ is the abbreviation for the cosine of angle $A$. Cosecant is abbreviated as $\csc A$ or $\text{cosec } A$.

(iv) $\cot A$ is the product of $\cot$ and $A$.

False. ‘$\cot$‘ separated from $A$ has no meaning. It denotes the cotangent function calculated for angle $A$.

(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.

False. $\sin \theta = \frac{\text{side opposite}}{\text{hypotenuse}}$. Since the hypotenuse is always the longest side in a right-angled triangle, the value of $\sin \theta$ can never be greater than 1. Here $\frac{4}{3} \approx 1.33 > 1$.

Q12: In a triangle $ABC$, $\angle B = 90^\circ$, $\sin A = \frac{3}{5}$ then the value of $\cos A$ is:

(a) $\frac{5}{3}$

(b) $\frac{3}{5}$

(c) $\frac{4}{5}$

(d) $\frac{5}{4}$

Solution:

Given $\sin A = \frac{BC}{AC} = \frac{3}{5}$. Let $BC = 3k$ and $AC = 5k$.

By using the Pythagoras Theorem, we have:

$AB^2 = AC^2 – BC^2$

$\implies AB^2 = (5k)^2 – (3k)^2$

$\implies AB^2 = 25k^2 – 9k^2$

$\implies AB^2 = 16k^2$

$\implies AB = \sqrt{16k^2}$

$\therefore AB = 4k$

Therefore,

$\cos A = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Hence, the correct option is (c).