NCERT Exemplar Solutions Class 10 Maths Introduction to Trigonometry and its Applications — Exercise 8.1

NCERT Exemplar Problems, Class 10 Mathematics — Introduction to Trigonometry and its Applications, Exercise 8.1. All 15 multiple-choice questions solved.

Fifteen questions, and almost every one of them collapses in a line once you see which of four tools it wants. They are:

The Pythagorean identities, which turn any one ratio into any other:

$$\sin^2\theta + \cos^2\theta = 1, \qquad \sec^2\theta – \tan^2\theta = 1, \qquad \operatorname{cosec}^2\theta – \cot^2\theta = 1$$

The specific-angle table for $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, which you should be able to write out from memory in under a minute.

The complementary-angle relations, which swap a ratio for its co-ratio:

$$\sin(90^\circ – \theta) = \cos\theta, \qquad \tan(90^\circ – \theta) = \cot\theta, \qquad \sec(90^\circ – \theta) = \operatorname{cosec}\theta$$

The right triangle itself — given one ratio, draw the triangle, find the third side by Pythagoras, and read off whatever else is asked.

A note on the current syllabus. The complementary-angle relations are no longer in the Class 10 textbook. The rationalised chapter runs Introduction → Trigonometric Ratios → Ratios of Some Specific Angles → Trigonometric Identities → Summary, and that is all; the word “complementary” survives exactly once in the whole chapter, in a historical note explaining where the word cosine came from. The Summary’s six points do not include a single complementary-angle identity.

So Questions 3, 5, 6, 7, 11 and 14 here — six of the fifteen — test material the book no longer carries. Each of those is flagged where it appears. They are worth learning anyway: the relations are two lines to derive from a right triangle, they return in Class 11, and they are the difference between a one-line answer and a page of algebra.

Key insight. When an expression pairs two angles adding to $90^\circ$ — $22^\circ$ with $68^\circ$, $47^\circ$ with $43^\circ$, $75^\circ + \theta$ with $15^\circ – \theta$ — it has been built so that the two halves cancel or combine to $1$. Scan for that pairing before computing anything. Nine of these fifteen are decided by it.

Choose the correct answer from the given four options:

Question 1

If $\cos A = \dfrac{4}{5}$, then the value of $\tan A$ is

Solution. Draw the triangle rather than juggling identities. $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac45$, so take the side adjacent to $A$ as $4$ and the hypotenuse as $5$. By Pythagoras the opposite side is

$$\sqrt{5^2 – 4^2} = \sqrt{25 – 16} = 3$$

The famous $3$–$4$–$5$ triangle. Then

$$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$$

The distractors are the other three ratios from the same triangle: $\frac35$ is $\sin A$, $\frac43$ is $\cot A$, and $\frac53$ is $\operatorname{cosec} A$. Getting one of those means the triangle was right and the final reading was not — so label the sides “opposite”, “adjacent”, “hypotenuse” on the sketch rather than trusting which number goes where.

Question 2

If $\sin A = \dfrac{1}{2}$, then the value of $\cot A$ is

Solution. $\sin A = \frac12$ identifies the angle outright: $A = 30^\circ$. From the standard table, $\cot 30^\circ = \sqrt3$.

Or without naming the angle, from the triangle: opposite $= 1$, hypotenuse $= 2$, so adjacent $= \sqrt{4-1} = \sqrt3$, and

$$\cot A = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt3}{1} = \sqrt3$$

Option (B) is $\tan 30^\circ$ — the reciprocal, and the single commonest slip in the subject. $\cot$ is adjacent over opposite; $\tan$ is opposite over adjacent. Option (C) is $\cos 30^\circ$.

Question 3

The value of the expression $\left[\operatorname{cosec}(75^\circ + \theta) – \sec(15^\circ – \theta) – \tan(55^\circ + \theta) + \cot(35^\circ – \theta)\right]$ is

Solution. Add the angles in each pair and the design of the question appears:

$$(75^\circ + \theta) + (15^\circ – \theta) = 90^\circ, \qquad (55^\circ + \theta) + (35^\circ – \theta) = 90^\circ$$

The $\theta$ cancels in both. So the four terms are two complementary pairs, and each pair annihilates.

First pair. Since $\operatorname{cosec}(90^\circ – x) = \sec x$, put $x = 15^\circ – \theta$:

$$\operatorname{cosec}(75^\circ + \theta) = \operatorname{cosec}\big[90^\circ – (15^\circ – \theta)\big] = \sec(15^\circ – \theta)$$

so $\operatorname{cosec}(75^\circ + \theta) – \sec(15^\circ – \theta) = 0$.

Second pair. Since $\tan(90^\circ – x) = \cot x$, put $x = 35^\circ – \theta$:

$$\tan(55^\circ + \theta) = \tan\big[90^\circ – (35^\circ – \theta)\big] = \cot(35^\circ – \theta)$$

so $-\tan(55^\circ + \theta) + \cot(35^\circ – \theta) = 0$.

The whole expression is $0$, for every $\theta$.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 4

Given that $\sin\theta = \dfrac{a}{b}$, then $\cos\theta$ is equal to

Solution. Straight from the Pythagorean identity:

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

$$\cos\theta = \frac{\sqrt{b^2 – a^2}}{b}$$

taking the positive root because $\theta$ is acute.

From the triangle it is the same statement: opposite $= a$, hypotenuse $= b$, so adjacent $= \sqrt{b^2 – a^2}$, and $\cos\theta$ is adjacent over hypotenuse.

Notice that the hypotenuse $b$ stays in the denominator. Option (A) has it upstairs, which is $\sec\theta$; option (D) is $\tan\theta$; option (B) is $\operatorname{cosec}\theta$. Every distractor is a real ratio of the same triangle, so the reasoning has to end with a check of which ratio was asked for.

Question 5

If $\cos(\alpha + \beta) = 0$, then $\sin(\alpha – \beta)$ can be reduced to

Solution. $\cos(\alpha+\beta) = 0$ and $\alpha + \beta$ is in the range where cosine vanishes only at $90^\circ$, so

$$\alpha + \beta = 90^\circ \implies \alpha = 90^\circ – \beta$$

Substitute into the expression asked about:

$$\alpha – \beta = (90^\circ – \beta) – \beta = 90^\circ – 2\beta$$

$$\sin(\alpha – \beta) = \sin(90^\circ – 2\beta) = \cos 2\beta$$

The last step is the complementary-angle relation $\sin(90^\circ – x) = \cos x$ with $x = 2\beta$.

Why does substituting for $\alpha$ rather than $\beta$ get you there? Because the four options are written in terms of either $\beta$ or $\alpha$, and only one of the two substitutions produces something on the list. Eliminating $\alpha$ gives $\cos 2\beta$, which is option (B); eliminating $\beta$ instead gives $\sin(2\alpha – 90^\circ) = -\cos 2\alpha$, which is not offered. If your first substitution produces nothing recognisable, try the other one before assuming you have made a mistake.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 6

The value of $(\tan 1^\circ \tan 2^\circ \tan 3^\circ \ldots \tan 89^\circ)$ is

Solution. Eighty-nine factors, and not one of them is worth evaluating. Pair the first with the last, the second with the second-last, and so on:

$$\tan 1^\circ \cdot \tan 89^\circ, \qquad \tan 2^\circ \cdot \tan 88^\circ, \qquad \ldots, \qquad \tan 44^\circ \cdot \tan 46^\circ$$

Each pair has angles summing to $90^\circ$, and $\tan(90^\circ – x) = \cot x = \dfrac{1}{\tan x}$, so each pair multiplies to

$$\tan x \cdot \frac{1}{\tan x} = 1$$

That accounts for $88$ of the factors in $44$ pairs, each equal to $1$. The only one left over is the middle factor $\tan 45^\circ$, which is itself $1$.

So the product is $1^{44} \times 1 = 1$.

Answering $0$ is tempting because the list is long and one expects something to vanish, but no factor in the range $1^\circ$ to $89^\circ$ is zero — $\tan 0^\circ$ would be, and it is not in the list.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 7

If $\cos 9\alpha = \sin\alpha$ and $9\alpha < 90^\circ$, then the value of $\tan 5\alpha$ is

Solution. Rewrite one side so that both are the same ratio. Using $\sin\alpha = \cos(90^\circ – \alpha)$:

$$\cos 9\alpha = \cos(90^\circ – \alpha)$$

The condition $9\alpha < 90^\circ$ is there precisely so that both angles are acute, and cosine takes each value only once on that range. So the angles themselves are equal:

$$9\alpha = 90^\circ – \alpha \implies 10\alpha = 90^\circ \implies \alpha = 9^\circ$$

Therefore $5\alpha = 45^\circ$ and

$$\tan 5\alpha = \tan 45^\circ = 1$$

The question is built so that $5\alpha$ lands exactly on $45^\circ$ — the one angle whose tangent is $1$ — which is a good check that $\alpha = 9^\circ$ is right. And $9\alpha = 81^\circ < 90^\circ$, so the stated condition is satisfied.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 8

If $\triangle ABC$ is right angled at $C$, then the value of $\cos(A + B)$ is

Solution. The three angles of a triangle sum to $180^\circ$, and $C = 90^\circ$, so

$$A + B = 180^\circ – 90^\circ = 90^\circ$$

Therefore

$$\cos(A + B) = \cos 90^\circ = 0$$

That is the whole question — no expansion of $\cos(A+B)$ is needed, and indeed the addition formula is not Class 10 material. The work is entirely in noticing that $A + B$ is a single known angle.

This one, unlike Questions 3, 5, 6 and 7, needs nothing off the current syllabus: the angle sum of a triangle is Class 9 and $\cos 90^\circ = 0$ is in the specific-angle table.

Question 9

If $\sin A + \sin^2 A = 1$, then the value of the expression $(\cos^2 A + \cos^4 A)$ is

Solution. Rearrange the given condition:

$$\sin A = 1 – \sin^2 A = \cos^2 A$$

That single substitution is the whole problem. Now rewrite the target using it:

$$\cos^2 A + \cos^4 A = \cos^2 A + \left(\cos^2 A\right)^2 = \sin A + \sin^2 A$$

And $\sin A + \sin^2 A = 1$ is exactly what was given. So the expression equals $1$.

The elegance is worth pausing on: the condition says that $\sin A$ and $\cos^2 A$ are the same number, so any expression in one can be rewritten in the other. The target was built to turn into the condition.

You could instead solve for $\sin A$ numerically — $\sin^2 A + \sin A – 1 = 0$ gives $\sin A = \frac{\sqrt5 – 1}{2} \approx 0.618$ — and grind it out. It works, but it takes ten times as long and invites arithmetic error.

Question 10

Given that $\sin\alpha = \dfrac{1}{2}$ and $\cos\beta = \dfrac{1}{2}$, then the value of $(\alpha + \beta)$ is

Solution. Read each angle off the table. $\sin\alpha = \frac12$ gives $\alpha = 30^\circ$; $\cos\beta = \frac12$ gives $\beta = 60^\circ$. So

$$\alpha + \beta = 30^\circ + 60^\circ = 90^\circ$$

The trap is reading both as $30^\circ$, which gives $60^\circ$ and option (C). Keep the two straight: sine is $\frac12$ at $30^\circ$, cosine is $\frac12$ at $60^\circ$. Sine climbs from $0$ to $1$ as the angle grows; cosine falls from $1$ to $0$. So the small value belongs to the small angle for sine and to the large angle for cosine.

Question 11

The value of the expression $\left[\dfrac{\sin^2 22^\circ + \sin^2 68^\circ}{\cos^2 22^\circ + \cos^2 68^\circ} + \sin^2 63^\circ + \cos 63^\circ \sin 27^\circ\right]$ is

Solution. Three complementary pairs are hidden here: $22^\circ$ with $68^\circ$, and $63^\circ$ with $27^\circ$. Convert everything to one angle per pair.

The numerator. $\sin 68^\circ = \sin(90^\circ – 22^\circ) = \cos 22^\circ$, so

$$\sin^2 22^\circ + \sin^2 68^\circ = \sin^2 22^\circ + \cos^2 22^\circ = 1$$

The denominator. $\cos 68^\circ = \cos(90^\circ – 22^\circ) = \sin 22^\circ$, so

$$\cos^2 22^\circ + \cos^2 68^\circ = \cos^2 22^\circ + \sin^2 22^\circ = 1$$

The fraction is therefore $\dfrac{1}{1} = 1$.

The remaining terms. $\sin 27^\circ = \sin(90^\circ – 63^\circ) = \cos 63^\circ$, so

$$\cos 63^\circ \sin 27^\circ = \cos 63^\circ \cdot \cos 63^\circ = \cos^2 63^\circ$$

and then

$$\sin^2 63^\circ + \cos^2 63^\circ = 1$$

Total: $1 + 1 = 2$.

Every pair was engineered to give exactly $1$, which is the shape of almost every question in this exercise. If a pairing is leaving you with a stray angle, the pairing is wrong.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 12

If $4\tan\theta = 3$, then $\left(\dfrac{4\sin\theta – \cos\theta}{4\sin\theta + \cos\theta}\right)$ is equal to

Solution. The expression is homogeneous — every term is degree one in the ratios — so dividing top and bottom by $\cos\theta$ turns it entirely into tangents:

$$\frac{4\sin\theta – \cos\theta}{4\sin\theta + \cos\theta} = \frac{\dfrac{4\sin\theta}{\cos\theta} – 1}{\dfrac{4\sin\theta}{\cos\theta} + 1} = \frac{4\tan\theta – 1}{4\tan\theta + 1}$$

Now $4\tan\theta = 3$ can be substituted as a block, without ever finding $\tan\theta$ itself:

$$= \frac{3 – 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$$

Dividing by $\cos\theta$ is the standard move for any expression where numerator and denominator have the same total degree in $\sin$ and $\cos$. It works because the division cancels in the fraction, so nothing is assumed about $\cos\theta$ beyond its being non-zero — true here, since $\theta$ is acute.

Doing it the long way — $\tan\theta = \frac34$, so $\sin\theta = \frac35$ and $\cos\theta = \frac45$ — gives $\frac{12/5 – 4/5}{12/5 + 4/5} = \frac{8}{16} = \frac12$. The same answer, three times the work.

Question 13

If $\sin\theta – \cos\theta = 0$, then the value of $(\sin^4\theta + \cos^4\theta)$ is

Solution. $\sin\theta = \cos\theta$ means $\tan\theta = 1$, so $\theta = 45^\circ$ and both ratios equal $\frac{1}{\sqrt2}$. Then

$$\sin^4\theta + \cos^4\theta = \left(\frac{1}{\sqrt2}\right)^4 + \left(\frac{1}{\sqrt2}\right)^4 = \frac14 + \frac14 = \frac12$$

The identity route is worth knowing because it generalises. Since $\sin^2\theta + \cos^2\theta = 1$, squaring gives

$$\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta = 1 \implies \sin^4\theta + \cos^4\theta = 1 – 2\sin^2\theta\cos^2\theta$$

With $\sin^2\theta = \cos^2\theta = \frac12$ this is $1 – 2 \cdot \frac12 \cdot \frac12 = 1 – \frac12 = \frac12$.

Answering $1$ means treating $\sin^4 + \cos^4$ as though it were $\sin^2 + \cos^2$. Fourth powers are strictly smaller than squares for numbers between $0$ and $1$, so the answer must come out below $1$ — a useful sanity check before looking at the options.

Question 14

$\sin(45^\circ + \theta) – \cos(45^\circ – \theta)$ is equal to

Solution. The two angles sum to $90^\circ$:

$$(45^\circ + \theta) + (45^\circ – \theta) = 90^\circ$$

so they are complementary, and

$$\cos(45^\circ – \theta) = \cos\big[90^\circ – (45^\circ + \theta)\big] = \sin(45^\circ + \theta)$$

The expression is therefore $\sin(45^\circ + \theta) – \sin(45^\circ + \theta) = 0$, for every $\theta$.

Test it with a value if the manipulation feels slippery. At $\theta = 15^\circ$: $\sin 60^\circ – \cos 30^\circ = \frac{\sqrt3}{2} – \frac{\sqrt3}{2} = 0$. At $\theta = 0^\circ$: $\sin 45^\circ – \cos 45^\circ = 0$. Substituting a convenient value is a legitimate way to kill three of four options in a multiple-choice question, and $\theta = 0^\circ$ eliminates (A), (C) and (D) at once.

This question uses the complementary-angle relations, which are not in the current Class 10 textbook.

Question 15

A pole $6$ m high casts a shadow $2\sqrt{3}$ m long on the ground, then the Sun’s elevation is

Solution. The pole is vertical and the shadow lies along the ground, so they meet at a right angle. The Sun’s elevation $\theta$ is the angle at the far tip of the shadow, between the ground and the line from there to the top of the pole.

Relative to $\theta$, the pole is the opposite side and the shadow is the adjacent side, so the ratio to use is the tangent:

$$\tan\theta = \frac{\text{height of pole}}{\text{length of shadow}} = \frac{6}{2\sqrt3} = \frac{3}{\sqrt3} = \sqrt3$$

simplifying because $\frac{3}{\sqrt3} = \frac{3}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{3\sqrt3}{3} = \sqrt3$.

And $\tan\theta = \sqrt3$ gives $\theta = 60^\circ$.

A check that needs no arithmetic: the shadow ($2\sqrt3 \approx 3.46$ m) is shorter than the pole ($6$ m), so the Sun is high in the sky and the elevation must exceed $45^\circ$. That rules out (B) and (C) immediately, and (D) would mean the Sun directly overhead with no shadow at all.

Common mistakes

• Questions 1, 2 and 4 — finding the triangle and then reading the wrong ratio off it. Every distractor in those three is a genuine ratio of the correct triangle. Label the sketch “opposite / adjacent / hypotenuse” and re-read the question before choosing.
• Question 2 — confusing $\tan$ with $\cot$. $\tan\theta$ is opposite over adjacent and $\cot\theta$ is its reciprocal. The pairing of a ratio with its co-ratio ($\sin$/$\operatorname{cosec}$, $\cos$/$\sec$, $\tan$/$\cot$) is worth over-learning, because the whole of this exercise leans on it.
• Question 10 — reading $\frac12$ as $30^\circ$ for both ratios. $\sin 30^\circ = \frac12$ but $\cos 60^\circ = \frac12$. Sine increases with the angle and cosine decreases, so the same value sits at complementary angles for the two.
• Question 13 — treating $\sin^4\theta + \cos^4\theta$ as $1$. Only the sum of squares is $1$. For any acute angle the fourth powers are smaller, so the value is always strictly less than $1$; in fact $\sin^4\theta + \cos^4\theta = 1 – 2\sin^2\theta\cos^2\theta$.
• Questions 3, 6, 11 and 14 — computing instead of pairing. Adding the two angles in each term takes five seconds and reveals whether the pair is complementary. No value of $\tan 37^\circ$ is ever needed in this exercise; if you find yourself wanting one, look for the pairing you have missed.
• Question 9 — solving the quadratic. $\sin A + \sin^2 A = 1$ rearranges to $\sin A = \cos^2 A$, which is a substitution rule, not an equation to solve. When a condition can be read as “these two expressions are the same thing”, use it to rewrite the target rather than to find a number.
• Question 12 — finding $\sin\theta$ and $\cos\theta$ separately. The expression is homogeneous, so dividing through by $\cos\theta$ reduces it to tangents and the given block $4\tan\theta = 3$ substitutes whole. Look for homogeneity before computing individual ratios.
• Question 15 — using $\sin$ or $\cos$ for a height-and-shadow problem. The pole and the shadow are the two legs of the right angle, never the hypotenuse, so the ratio connecting them is always the tangent. The hypotenuse here is the Sun’s ray, and no question asks for it.

Practise next

• Exercise 8.2 — twelve true-or-false statements, several of which are the same complementary-angle pairings dressed as claims to be judged. The justification is the marked part.
• Exercise 8.3 — fifteen short-answer questions: seven identities to prove, and eight numerical ones including the first height-and-distance problems.
• Exercise 7.1 — worth revisiting for the distance formula, which is Pythagoras in coordinates and is the same relation underlying every triangle in this unit.