NCERT Exemplar Solutions Class 10 Maths Introduction to Trigonometry and its Applications — Exercise 8.2

NCERT Exemplar Problems, Class 10 Mathematics — Introduction to Trigonometry and its Applications, Exercise 8.2. All 12 true-or-false questions solved.

Twelve statements to be judged, and the instruction is justify your answer — so the verdict is worth nothing on its own. What this exercise really tests is whether you can produce the one line that settles each claim.

Those lines come in four kinds here, and recognising which kind a statement wants is most of the work:

• an identity to be checked — expand both sides and see whether the difference is zero (Questions 4, 5, 6)
• a complementary pairing — do the two angles sum to $90^\circ$? (Questions 1, 2)
• a bound being violated — $\sin\theta$ and $\cos\theta$ never exceed $1$, and some expressions never drop below $2$ (Questions 9, 10)
• a physical claim about elevation — which usually turns on the fact that $\tan\theta$ increases with $\theta$, so a longer shadow means a lower Sun (Questions 7, 8, 11, 12)

A note on the current syllabus. The section “Trigonometric Ratios of Complementary Angles” has been removed from the Class 10 textbook. The chapter now runs Introduction → Trigonometric Ratios → Ratios of Some Specific Angles → Trigonometric Identities → Summary, and the word “complementary” appears exactly once in it, inside a historical note on where the word cosine came from. Questions 1 and 2 here rest squarely on those relations, and Question 3 is usually done with them — though it can be settled without, as shown below.

Key insight. Four of these twelve are false because a quantity is being asked to leave its permitted range. $\sin\theta$ and $\cos\theta$ live in $[-1, 1]$; $a + \frac1a$ is at least $2$ for positive $a$; $\frac{a^2+b^2}{2ab}$ is at least $1$. Before manipulating, ask whether the statement is even arithmetically possible.

Write ‘True’ or ‘False’ and justify your answer in each of the following:

Question 1

$\dfrac{\tan 47^\circ}{\cot 43^\circ} = 1$

Solution. The angles $47^\circ$ and $43^\circ$ sum to $90^\circ$, so they are complementary and one ratio converts into the other:

$$\cot 43^\circ = \cot(90^\circ – 47^\circ) = \tan 47^\circ$$

Therefore

$$\frac{\tan 47^\circ}{\cot 43^\circ} = \frac{\tan 47^\circ}{\tan 47^\circ} = 1$$

The numerator and denominator are literally the same number — approximately $1.0724$ — so the quotient is $1$ exactly, not merely close to it.

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

Question 2

The value of the expression $(\cos^2 23^\circ – \sin^2 67^\circ)$ is positive.

Solution. Again the angles sum to $90^\circ$: $23^\circ + 67^\circ = 90^\circ$. So

$$\sin 67^\circ = \sin(90^\circ – 23^\circ) = \cos 23^\circ$$

and the expression becomes

$$\cos^2 23^\circ – \cos^2 23^\circ = 0$$

Zero is not positive, so the statement is false.

This is a sharper trap than it looks. Both terms are individually positive — $\cos^2 23^\circ \approx 0.847$ and $\sin^2 67^\circ \approx 0.847$ — and a quick glance at “$\cos$ of a small angle minus $\sin$ of a large angle” might suggest a positive difference. They are equal, and the difference is exactly zero.

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

Question 3

The value of the expression $(\sin 80^\circ – \cos 80^\circ)$ is negative.

Solution. Convert one term so that both are the same ratio. Since $\cos 80^\circ = \sin(90^\circ – 80^\circ) = \sin 10^\circ$, the expression is

$$\sin 80^\circ – \sin 10^\circ$$

On the range $0^\circ$ to $90^\circ$ the sine increases with the angle, and $80^\circ > 10^\circ$, so $\sin 80^\circ > \sin 10^\circ$ and the difference is positive. The statement is false.

There is a route that avoids complementary angles altogether, and it is worth knowing since those relations are off the current syllabus. As the angle grows from $0^\circ$ to $90^\circ$, $\sin\theta$ rises from $0$ to $1$ while $\cos\theta$ falls from $1$ to $0$. They cross at $45^\circ$, and beyond that crossing the sine is the larger of the two. Since $80^\circ > 45^\circ$, we have $\sin 80^\circ > \cos 80^\circ$ and the difference is positive.

Numerically, $\sin 80^\circ \approx 0.985$ and $\cos 80^\circ \approx 0.174$, so the difference is about $0.81$ — not marginal at all.

Question 4

$\sqrt{(1 – \cos^2\theta)\sec^2\theta} = \tan\theta$

Solution. Work inside the root first. The Pythagorean identity gives $1 – \cos^2\theta = \sin^2\theta$, and $\sec^2\theta = \dfrac{1}{\cos^2\theta}$, so

$$(1 – \cos^2\theta)\sec^2\theta = \sin^2\theta \cdot \frac{1}{\cos^2\theta} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$$

Taking the square root:

$$\sqrt{\tan^2\theta} = |\tan\theta|$$

and for an acute $\theta$ the tangent is positive, so $|\tan\theta| = \tan\theta$. The statement is true.

The modulus matters, and it is why this is worth more than a line. $\sqrt{x^2}$ is $|x|$, not $x$ — a square root is by convention the non-negative root. The identity as printed is true because Class 10 restricts $\theta$ to acute angles, where $\tan\theta \ge 0$. Written for a general angle it would have to read $\sqrt{(1-\cos^2\theta)\sec^2\theta} = |\tan\theta|$, and at, say, $\theta = 120^\circ$ the left side is $\sqrt3$ while $\tan\theta$ is $-\sqrt3$.

The textbook’s own restriction is explicit: “We will restrict our discussion to acute angles only.”

Question 5

If $\cos A + \cos^2 A = 1$, then $\sin^2 A + \sin^4 A = 1$.

Solution. Rearrange what is given:

$$\cos A = 1 – \cos^2 A = \sin^2 A$$

So the condition says that $\cos A$ and $\sin^2 A$ are the same number. That is a substitution rule, and it turns the target into the condition:

$$\sin^2 A + \sin^4 A = \sin^2 A + \left(\sin^2 A\right)^2 = \cos A + \cos^2 A = 1$$

True, and the proof is three lines with no quadratic solved anywhere.

This is the same architecture as Question 9 of Exercise 8.1, where $\sin A + \sin^2 A = 1$ forced $\sin A = \cos^2 A$ and the target $\cos^2 A + \cos^4 A$ collapsed the same way. Once you see a condition of the form “ratio $=$ square of the other ratio”, look for the target that is its mirror image.

Question 6

$(\tan\theta + 2)(2\tan\theta + 1) = 5\tan\theta + \sec^2\theta$

Solution. Expand the left side:

$$(\tan\theta + 2)(2\tan\theta + 1) = 2\tan^2\theta + \tan\theta + 4\tan\theta + 2 = 2\tan^2\theta + 5\tan\theta + 2$$

Now the right side, using $\sec^2\theta = 1 + \tan^2\theta$:

$$5\tan\theta + \sec^2\theta = 5\tan\theta + 1 + \tan^2\theta = \tan^2\theta + 5\tan\theta + 1$$

Subtract:

$$\text{LHS} – \text{RHS} = \left(2\tan^2\theta + 5\tan\theta + 2\right) – \left(\tan^2\theta + 5\tan\theta + 1\right) = \tan^2\theta + 1 = \sec^2\theta$$

The difference is $\sec^2\theta$, which is at least $1$ for every angle and is never zero. So the two sides never agree, for any $\theta$ whatsoever — this is not an identity that fails at a few awkward angles, it fails everywhere.

That is the best kind of justification for a false statement: not a counterexample but a proof that no example can exist. A single counterexample would do for marks — at $\theta = 45^\circ$ the left side is $3 \times 3 = 9$ and the right side is $5 + 2 = 7$ — but naming the gap as $\sec^2\theta$ says more.

Question 7

If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

Solution. For a tower of fixed height $h$ casting a shadow of length $s$, the Sun’s elevation $\theta$ satisfies

$$\tan\theta = \frac{h}{s}$$

Since $h$ is fixed, a larger $s$ makes the fraction smaller, so $\tan\theta$ decreases. And on $0^\circ$ to $90^\circ$ the tangent increases with the angle, so a smaller tangent means a smaller angle. The elevation decreases.

The statement claims the opposite, so it is false.

Everyday experience agrees. Shadows are shortest near midday, when the Sun is highest, and they lengthen through the afternoon as the Sun drops towards the horizon. A long shadow means a low Sun.

A concrete check: a $10$ m tower with a $10$ m shadow gives $\tan\theta = 1$, so $\theta = 45^\circ$. Lengthen the shadow to $10\sqrt3$ m and $\tan\theta = \frac{1}{\sqrt3}$, so $\theta = 30^\circ$. The shadow grew and the angle fell.

Question 8

If a man standing on a platform $3$ metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Solution. Let the cloud be at height $H$ above the lake surface, and let the horizontal distance from the man to the point directly below the cloud be $d$.

The reflection is the mirror image of the cloud in the water surface, so it lies $H$ below the surface.

The man’s eye is $3$ m above the surface, and both angles are measured from the horizontal through his eye. So:

• the cloud is $H – 3$ metres above eye level, giving $\ \tan\alpha = \dfrac{H – 3}{d}$
• the reflection is $H + 3$ metres below eye level, giving $\ \tan\beta = \dfrac{H + 3}{d}$

Since $H + 3 > H – 3$, we have $\tan\beta > \tan\alpha$ and therefore $\beta > \alpha$. The angle of depression of the reflection is strictly greater than the angle of elevation of the cloud, so the statement is false.

Why the platform is the whole question. If the observer’s eye were exactly at water level, the two vertical distances would both be $H$ and the angles would be equal — the water surface would be the mirror line and everything would be symmetric about the line of sight. The $3$ metres of platform breaks that symmetry, and it breaks it by $6$ metres in total: the reflection is displaced $3$ m further from eye level in one direction while the cloud is $3$ m nearer in the other.

Question 9

The value of $2\sin\theta$ can be $a + \dfrac{1}{a}$, where $a$ is a positive number, and $a \ne 1$.

Solution. Two ranges have to be compared, and they do not overlap.

The left side. For any angle, $\sin\theta \le 1$, so $2\sin\theta \le 2$.

The right side. For positive $a$,

$$a + \frac{1}{a} – 2 = \frac{a^2 – 2a + 1}{a} = \frac{(a-1)^2}{a}$$

The numerator is a square, so it is non-negative, and the denominator is positive. The fraction is therefore non-negative, giving $a + \frac1a \ge 2$ — with equality only when $a = 1$. Since the question excludes $a = 1$, we have strictly

$$a + \frac{1}{a} > 2$$

So the right side is always greater than $2$ and the left side never exceeds $2$. They can never be equal, and the statement is false.

Try it and see: $a = 2$ gives $2.5$, $a = \frac12$ gives $2.5$, $a = 3$ gives $3.33$, $a = 1.1$ gives $2.009$. Always above $2$, and $2\sin\theta$ never gets there except at $\theta = 90^\circ$ where it equals $2$ exactly — which the exclusion $a \ne 1$ rules out on the other side.

Question 10

$\cos\theta = \dfrac{a^2 + b^2}{2ab}$, where $a$ and $b$ are two distinct numbers such that $ab > 0$.

Solution. The same argument as Question 9, one step more general.

$$\frac{a^2+b^2}{2ab} – 1 = \frac{a^2 – 2ab + b^2}{2ab} = \frac{(a-b)^2}{2ab}$$

Since $a \ne b$, the numerator $(a-b)^2$ is strictly positive; since $ab > 0$, the denominator is positive. So the fraction is strictly positive and

$$\frac{a^2+b^2}{2ab} > 1$$

But $\cos\theta \le 1$ for every angle. No angle has a cosine greater than $1$, so the equation has no solution and the statement is false.

The two conditions attached to the question are doing real work, and it is worth seeing what each rules out. Without $a \ne b$ the expression could equal $1$ exactly, and $\theta = 0^\circ$ would work. Without $ab > 0$ — say $a = 1$, $b = -1$ — the expression would be $\frac{2}{-2} = -1$, and $\theta = 180^\circ$ would satisfy it. The question has been set up so that neither escape is available.

Question 11

The angle of elevation of the top of a tower is $30^\circ$. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

Solution. Let the observer stand a fixed distance $d$ from the foot of a tower of height $h$. Then

$$\tan 30^\circ = \frac{h}{d} \implies \frac{h}{d} = \frac{1}{\sqrt3}$$

Doubling the height to $2h$ while keeping $d$ the same gives a new elevation $\theta’$ with

$$\tan\theta’ = \frac{2h}{d} = 2 \times \frac{1}{\sqrt3} = \frac{2}{\sqrt3} \approx 1.1547$$

If the angle had doubled to $60^\circ$, the tangent would be $\tan 60^\circ = \sqrt3 \approx 1.732$. It is not: $\frac{2}{\sqrt3} \ne \sqrt3$, and in fact $\theta’ = \arctan(1.1547) \approx 49.1^\circ$.

So doubling the height raises the elevation from $30^\circ$ to about $49^\circ$ — a substantial increase, but well short of double.

The general reason is worth more than this instance. Doubling the height doubles the tangent, not the angle, and the tangent is not proportional to the angle. It grows slowly at first and then explosively: $\tan 30^\circ \approx 0.58$, $\tan 60^\circ \approx 1.73$ — three times as much for double the angle, not twice. Any claim that scaling a length scales the angle by the same factor is false for the same reason.

Question 12

If the height of a tower and the distance of the point of observation from its foot, both, are increased by $10\%$, then the angle of elevation of its top remains unchanged.

Solution. Let the original height be $h$ and the original distance $d$, so that the elevation $\theta$ satisfies

$$\tan\theta = \frac{h}{d}$$

Increasing each by $10\%$ multiplies both by $1.1$. The new elevation $\theta’$ satisfies

$$\tan\theta’ = \frac{1.1h}{1.1d} = \frac{h}{d} = \tan\theta$$

The factor cancels, so $\tan\theta’ = \tan\theta$ and hence $\theta’ = \theta$. The angle is unchanged, and the statement is true.

What makes this work — and what makes Question 11 fail — is that the angle depends only on the ratio of the two lengths, never on their individual sizes. Scaling both by the same factor leaves the ratio alone, so the new triangle is similar to the old one and all its angles are identical. Scaling only one of them, as in Question 11, changes the ratio and therefore the angle.

Nothing about $10\%$ is special. Any common factor would do, and the same argument shows that a tower twice as tall seen from twice the distance subtends exactly the same angle.

Common mistakes

• Question 4 — dropping the modulus after a square root. $\sqrt{x^2} = |x|$, and the printed identity is true only because Class 10 restricts $\theta$ to acute angles where $\tan\theta \ge 0$. Say so; it is the difference between a correct statement and a lucky one.
• Questions 9 and 10 — trying to solve the equation. Neither has a solution, and hunting for one wastes the whole question. Compare the ranges instead: if one side never drops below $2$ and the other never rises above $2$, there is nothing more to say.
• Question 6 — quoting a counterexample and stopping. A single angle where the two sides differ does answer the question, but computing LHS $-$ RHS $= \sec^2\theta$ proves that no angle works, and takes the same three lines.
• Questions 7 and 11 — treating the angle as proportional to the length. It is the tangent that scales with the height, and the tangent is a badly non-linear function of the angle. Doubling a height never doubles an elevation.
• Question 8 — forgetting that the eye is not at water level. The reflection lies $H$ below the surface, which is $H + 3$ below eye level, while the cloud is only $H – 3$ above it. Equality would need the observer’s eye exactly at water level.
• Questions 1, 2 and 3 — reaching for a calculator. Every one of them is settled by noticing that the two angles sum to $90^\circ$. Add the angles first; the arithmetic is almost never required.
• Question 12 — thinking a percentage change must change something. The angle depends only on the ratio $\frac{h}{d}$, and a common factor cancels out of a ratio. That is exactly the condition for similar triangles, which preserve every angle.

Practise next

• Exercise 8.3 — fifteen short-answer questions: seven identities to prove and eight numerical ones, including the first height-and-distance problems of the unit.
• Exercise 8.4 — eighteen long-answer questions, where the identities of Questions 4 to 6 here reappear in harder disguises and the elevation reasoning of Questions 7 to 12 becomes full two-position problems.
• Exercise 8.1 — worth a second pass for the complementary-angle pairings, which decide six of its fifteen questions and three of these twelve.