NCERT Class 10 Mathematics — Introduction to Trigonometry, Exercise 8.1. All 11 questions solved.

Exercise 8.1 is the definitions exercise of the chapter. Every question is answered from one right triangle and the six ratios defined on it. In a triangle $ABC$ right-angled at $B$, measured from the acute angle $A$:

$$\sin A = \frac{\text{side opposite } A}{\text{hypotenuse}} = \frac{BC}{AC}, \qquad \cos A = \frac{\text{side adjacent to } A}{\text{hypotenuse}} = \frac{AB}{AC}, \qquad \tan A = \frac{BC}{AB}$$

and the other three are their reciprocals:

$$\operatorname{cosec} A = \frac{1}{\sin A}, \qquad \sec A = \frac{1}{\cos A}, \qquad \cot A = \frac{1}{\tan A}$$

The only other tool needed is Pythagoras’ theorem, $AC^2 = AB^2 + BC^2$, to find whichever side the question leaves out.

Key insight. A trigonometric ratio fixes the shape of a right triangle, not its size. If $\sin A = \frac34$, the side opposite $A$ and the hypotenuse can be taken as $3k$ and $4k$ for any positive $k$, and $k$ cancels out of every ratio you then compute. So questions 3 to 10 all run on one routine: turn the given ratio into two sides, find the third by Pythagoras, and read off what is asked. The one thing to watch is that “opposite” and “adjacent” are always relative to the angle in hand — the same side is opposite $A$ and adjacent to $C$.

Question 1

In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine:

(i) $\sin A$, $\cos A$    (ii) $\sin C$, $\cos C$

Solution. Every one of these ratios has the hypotenuse $AC$ in its denominator, and the question does not give it, so find it first. The right angle is at $B$, which makes $AC$ the hypotenuse:

$$AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625 \quad\Longrightarrow\quad AC = 25 \text{ cm}$$

(i) Seen from $A$, the opposite side is $BC$ and the adjacent side is $AB$:

$$\sin A = \frac{BC}{AC} = \frac{7}{25}, \qquad \cos A = \frac{AB}{AC} = \frac{24}{25}$$

(ii) Seen from $C$, the roles swap: $AB$ is now opposite and $BC$ is adjacent.

$$\sin C = \frac{AB}{AC} = \frac{24}{25}, \qquad \cos C = \frac{BC}{AC} = \frac{7}{25}$$

Notice that $\sin A = \cos C$ and $\cos A = \sin C$. That is no coincidence: $A$ and $C$ are the two acute angles of one right triangle, so the side opposite one of them is the side adjacent to the other.

Question 2

In Fig. 8.13, find $\tan P – \cot R$.

Solution. The figure gives $PQ = 12$ cm and the hypotenuse $PR = 13$ cm, with the right angle at $Q$. Both ratios asked for use the leg $QR$, so Pythagoras first:

$$QR = \sqrt{PR^2 – PQ^2} = \sqrt{169 – 144} = \sqrt{25} = 5 \text{ cm}$$

For angle $P$, the opposite side is $QR$ and the adjacent side is $PQ$:

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

For angle $R$, the adjacent side is $QR$ and the opposite side is $PQ$, and cotangent is adjacent over opposite:

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

The two are the same fraction, so

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

As in question 1, this happens because $P$ and $R$ are the two acute angles of one right triangle: the tangent of one is always the cotangent of the other.

Question 3

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

Solution. Draw a triangle $ABC$ right-angled at $B$. Since $\sin A = \dfrac{BC}{AC} = \dfrac34$, take $BC = 3k$ and $AC = 4k$ for some positive $k$. The missing leg is

$$AB = \sqrt{AC^2 – BC^2} = \sqrt{16k^2 – 9k^2} = \sqrt{7}\,k$$

Now read off the two ratios; $k$ cancels in each, which is why its value never matters:

$$\cos A = \frac{AB}{AC} = \frac{\sqrt7\,k}{4k} = \frac{\sqrt7}{4}, \qquad \tan A = \frac{BC}{AB} = \frac{3k}{\sqrt7\,k} = \frac{3}{\sqrt7}$$

Question 4

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

Solution. First isolate the ratio by dividing by $15$: $\cot A = \dfrac{8}{15}$. In $\triangle ABC$ right-angled at $B$, $\cot A = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{AB}{BC}$, so take $AB = 8k$ and $BC = 15k$. The hypotenuse is

$$AC = \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$$

Therefore

$$\sin A = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}, \qquad \sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{AC}{AB} = \frac{17}{8}$$

Question 5

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

Solution. $\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}$, so in a right triangle with acute angle $\theta$ take the hypotenuse as $13k$ and the side adjacent to $\theta$ as $12k$. The side opposite $\theta$ is then

$$\sqrt{(13k)^2 – (12k)^2} = \sqrt{169k^2 – 144k^2} = \sqrt{25k^2} = 5k$$

With opposite $5k$, adjacent $12k$ and hypotenuse $13k$, each ratio is read straight off its definition:

$$\sin\theta = \frac{5}{13}, \quad \cos\theta = \frac{12}{13}, \quad \tan\theta = \frac{5}{12}, \quad \cot\theta = \frac{12}{5}, \quad \operatorname{cosec}\theta = \frac{13}{5}$$

A quick check: $\cos\theta$ must be the reciprocal of the given $\sec\theta$, and $\dfrac{12}{13}$ is.

Question 6

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

Solution. Nothing says the two angles belong to the same triangle, so give each its own right triangle and make the two triangles as alike as the data allows.

Draw a right triangle $PQR$ with the right angle at $Q$ and $\angle P = \angle A$, and a second right triangle $XYZ$ with the right angle at $Y$ and $\angle X = \angle B$. A ratio does not depend on the size of the triangle, so we are free to scale the second triangle until the two hypotenuses are equal: $PR = XZ$.

Now use the given condition:

$$\cos A = \frac{PQ}{PR}, \qquad \cos B = \frac{XY}{XZ}, \qquad \cos A = \cos B \;\Longrightarrow\; \frac{PQ}{PR} = \frac{XY}{XZ}$$

and since $PR = XZ$, the numerators must be equal too: $PQ = XY$.

So the two right triangles have equal hypotenuses ($PR = XZ$) and a pair of equal legs ($PQ = XY$). By the RHS congruence rule, $\triangle PQR \cong \triangle XYZ$, and corresponding angles of congruent triangles are equal:

$$\angle P = \angle X, \qquad\text{that is,}\qquad \angle A = \angle B$$

Question 7

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

(i) $\dfrac{(1 + \sin\theta)(1 – \sin\theta)}{(1 + \cos\theta)(1 – \cos\theta)}$    (ii) $\cot^2\theta$

Solution. $\cot\theta = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac78$, so take the adjacent side as $7k$ and the opposite side as $8k$. The hypotenuse is $\sqrt{49k^2 + 64k^2} = \sqrt{113}\,k$, which gives

$$\sin\theta = \frac{8}{\sqrt{113}}, \qquad \cos\theta = \frac{7}{\sqrt{113}}$$

(i) Each pair of brackets is a difference of squares, $(1 + a)(1 – a) = 1 – a^2$, so simplify before substituting and the surd never has to be handled:

$$\frac{1 – \sin^2\theta}{1 – \cos^2\theta} = \frac{1 – \frac{64}{113}}{1 – \frac{49}{113}} = \frac{\;\frac{49}{113}\;}{\;\frac{64}{113}\;} = \frac{49}{64}$$

(ii)

$$\cot^2\theta = \left(\frac78\right)^2 = \frac{49}{64}$$

The two answers agree for a reason. $1 – \sin^2\theta = \cos^2\theta$ and $1 – \cos^2\theta = \sin^2\theta$, so the expression in part (i) is really $\dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$. That identity is proved in Section 8.4, and Exercise 8.3 leans on it throughout.

Question 8

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

Solution. $\cot A = \dfrac43$, so $\tan A = \dfrac34$. In a right triangle with acute angle $A$, take the opposite side as $3k$ and the adjacent side as $4k$; the hypotenuse is $\sqrt{9k^2 + 16k^2} = 5k$, so $\sin A = \dfrac35$ and $\cos A = \dfrac45$.

“Check whether” means evaluate each side on its own and compare — not assume the equation is true and manipulate it.

Left-hand side:

$$\frac{1 – \tan^2 A}{1 + \tan^2 A} = \frac{1 – \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\;\frac{7}{16}\;}{\;\frac{25}{16}\;} = \frac{7}{25}$$

Right-hand side:

$$\cos^2 A – \sin^2 A = \frac{16}{25} – \frac{9}{25} = \frac{7}{25}$$

Both sides equal $\dfrac{7}{25}$, so the statement holds.

Question 9

In triangle $ABC$, right-angled at $B$, if $\tan A = \dfrac{1}{\sqrt3}$, find the value of:

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

Solution. $\tan A = \dfrac{BC}{AB} = \dfrac{1}{\sqrt3}$, so take $BC = k$ and $AB = \sqrt3\,k$. Then

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{3k^2 + k^2} = 2k$$

Angle $A$ has $BC$ opposite and $AB$ adjacent; angle $C$ has them the other way round:

$$\sin A = \frac{BC}{AC} = \frac12, \quad \cos A = \frac{AB}{AC} = \frac{\sqrt3}{2}, \qquad \sin C = \frac{AB}{AC} = \frac{\sqrt3}{2}, \quad \cos C = \frac{BC}{AC} = \frac12$$

(i)

$$\sin A \cos C + \cos A \sin C = \frac12 \cdot \frac12 + \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{2} = \frac14 + \frac34 = 1$$

(ii)

$$\cos A \cos C – \sin A \sin C = \frac{\sqrt3}{2} \cdot \frac12 – \frac12 \cdot \frac{\sqrt3}{2} = \frac{\sqrt3}{4} – \frac{\sqrt3}{4} = 0$$

The tidy answers are a preview of Class 11, where these two expressions are recognised as $\sin(A + C)$ and $\cos(A + C)$. In a right triangle $A + C = 90^\circ$, so they could only have come out as $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$.

Question 10

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

Solution. Only one side is given outright, but the condition $PR + QR = 25$ is enough to pin down the other two. Let $QR = x$ cm; then $PR = (25 – x)$ cm. The right angle is at $Q$, so $PR$ is the hypotenuse and Pythagoras gives

$$PR^2 = PQ^2 + QR^2 \quad\Longrightarrow\quad (25 – x)^2 = 25 + x^2$$

$$625 – 50x + x^2 = 25 + x^2 \quad\Longrightarrow\quad 50x = 600 \quad\Longrightarrow\quad x = 12$$

The $x^2$ terms cancel, which is why a single linear equation is enough. So $QR = 12$ cm and $PR = 25 – 12 = 13$ cm. (Check: $5^2 + 12^2 = 169 = 13^2$.)

From $P$, the opposite side is $QR$ and the adjacent side is $PQ$:

$$\sin P = \frac{QR}{PR} = \frac{12}{13}, \qquad \cos P = \frac{PQ}{PR} = \frac{5}{13}, \qquad \tan P = \frac{QR}{PQ} = \frac{12}{5}$$

Question 11

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

(i) The value of $\tan A$ is always less than $1$. (ii) $\sec A = \dfrac{12}{5}$ for some value of angle $A$. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$. (iv) $\cot A$ is the product of $\cot$ and $A$. (v) $\sin\theta = \dfrac43$ for some angle $\theta$.

Solution.

(i) False. $\tan A = \dfrac{\text{opposite}}{\text{adjacent}}$, and nothing stops the opposite leg being the longer of the two. Question 10 has already produced one: $\tan P = \dfrac{12}{5} > 1$.

(ii) True. $\sec A = \dfrac{\text{hypotenuse}}{\text{adjacent}}$, and the hypotenuse is always longer than either leg, so $\sec A$ can be any number greater than $1$ — which $\dfrac{12}{5}$ is. A right triangle with hypotenuse $12$, adjacent side $5$ and opposite side $\sqrt{144 – 25} = \sqrt{119}$ gives exactly this value.

(iii) False. $\cos A$ abbreviates the cosine of $A$. The cosecant is written $\operatorname{cosec} A$.

(iv) False. $\cot A$ is a single symbol meaning the cotangent of the angle $A$. “$\cot$” on its own has no value, so it cannot be multiplied by anything; the $A$ is the angle the ratio is taken of, not a factor.

(v) False. $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, and the hypotenuse is the longest side, so $\sin\theta$ can never exceed $1$. Since $\dfrac43 > 1$, no angle has this sine.

Common mistakes

• Question 1(ii), keeping the “opposite” side from part (i). The side opposite $A$ is $BC$, but the side opposite $C$ is $AB$. Writing $\sin C = \frac{7}{25}$ means the labels from part (i) were carried across; identify opposite and adjacent afresh every time the angle changes.
• Question 2, reading $\cot R$ as $\frac{PQ}{QR}$. Cotangent is adjacent over opposite, and relative to $R$ the adjacent leg is $QR$. The slip comes from still looking at the triangle from $P$, the angle just used for $\tan P$.
• Question 4, taking $\cot A = \frac{15}{8}$. $15\cot A = 8$ means $\cot A = \frac{8}{15}$: divide by $15$, don’t flip the fraction. A wrong start here inverts both answers.
• Question 6, putting $A$ and $B$ in the same right triangle. A very common “proof” draws one triangle with acute angles $A$ and $B$ and compares their adjacent sides. But the two acute angles of a right triangle always add up to $90^\circ$, so that argument quietly assumes $A + B = 90^\circ$, which the question never says. Two separate triangles, as above, avoid the hidden assumption.
• Question 7, taking the hypotenuse as $7 + 8 = 15$. It is $\sqrt{7^2 + 8^2} = \sqrt{113}$ — not a sum, and not a whole number. Simplifying part (i) to $\frac{1 – \sin^2\theta}{1 – \cos^2\theta}$ first means the surd cancels as soon as it is squared.
• Question 10, guessing that the triangle is 5–12–13. It is, but a guess is not a solution. Set $QR = x$ and let Pythagoras produce $x = 12$.
• Question 11(ii), marking it false because “trigonometric ratios are less than 1”. That holds for $\sin$ and $\cos$ only. $\sec$ and $\operatorname{cosec}$ are never less than $1$, and $\tan$ and $\cot$ can take any positive value.

Practise next

• Exercise 8.2 — the ratios of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$, where the triangle of question 9 turns out to be the $30^\circ$–$60^\circ$–$90^\circ$ triangle.
• Exercise 8.3 — the identity $\sin^2 A + \cos^2 A = 1$ that explains why the two parts of question 7 agree.
• Class 11 Trigonometric Functions, Exercise 3.2 — finding the other five ratios from one, as in question 5, but for angles beyond $90^\circ$, where signs start to matter.