NCERT Exemplar Solutions Class 10 Maths Introduction to Trigonometry and its Applications — Exercise 8.3

NCERT Exemplar Problems, Class 10 Mathematics — Introduction to Trigonometry and its Applications, Exercise 8.3. All 15 questions solved.

Fifteen questions in two halves. Questions 1 to 7 are identities to prove; the rest are numerical, and Questions 8, 10 and 14 are the unit’s first genuine height-and-distance problems.

For the identities there is one method that handles nearly all of them, and it is worth stating as a procedure rather than a hope:

1. Work on one side only — usually the messier one — and drive it towards the other. Never manipulate both sides of a claim you have not yet proved.
2. Put everything over a common denominator when the side is a sum of fractions. Questions 1, 2 and 4 all fall to this in two lines.
3. Look for a difference of squares disguised as an identity. $\cot^2\alpha = \operatorname{cosec}^2\alpha – 1 = (\operatorname{cosec}\alpha – 1)(\operatorname{cosec}\alpha + 1)$ is exactly what Question 6 needs, and the same trick with $\sec^2\theta – 1 = \tan^2\theta$ settles Question 15.
4. Convert to $\sin$ and $\cos$ if nothing else presents itself. It always works; it is just usually slower.

A note on the current syllabus. The section “Trigonometric Ratios of Complementary Angles” has been removed from the Class 10 textbook, so Questions 7 and 13 test material the book no longer carries. Both are flagged where they appear.

Key insight. In an identity, the side with the fractions is the side to work on. Combining $\dfrac{\sin\theta}{1+\cos\theta} + \dfrac{1+\cos\theta}{\sin\theta}$ over one denominator produces $2 + 2\cos\theta$ on top, and the factor $(1 + \cos\theta)$ then cancels straight out. Nearly every identity here is built around one cancellation like that.

Prove the following (from Q.1 to Q.7):

Question 1

$$\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\operatorname{cosec}\theta$$

Solution. Take the left side and combine the two fractions over the common denominator $\sin\theta(1+\cos\theta)$:

$$\text{LHS} = \frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}$$

Expand the numerator, and use $\sin^2\theta + \cos^2\theta = 1$:

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

So

$$\text{LHS} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = 2\operatorname{cosec}\theta = \text{RHS}$$

The factor $(1+\cos\theta)$ cancelling is not luck — the identity was constructed so that it would, and spotting the $2 + 2\cos\theta$ as $2(1+\cos\theta)$ is the entire question.

Question 2

$$\frac{\tan A}{1 + \sec A} – \frac{\tan A}{1 – \sec A} = 2\operatorname{cosec} A$$

Solution. Combine over the common denominator $(1+\sec A)(1-\sec A)$, which is a difference of squares:

$$\text{LHS} = \tan A \cdot \frac{(1 – \sec A) – (1 + \sec A)}{(1+\sec A)(1-\sec A)} = \tan A \cdot \frac{-2\sec A}{1 – \sec^2 A}$$

Now $\sec^2 A – \tan^2 A = 1$ gives $1 – \sec^2 A = -\tan^2 A$, so

$$\text{LHS} = \frac{-2\tan A \sec A}{-\tan^2 A} = \frac{2\sec A}{\tan A}$$

Finally convert to sines and cosines:

$$\frac{2\sec A}{\tan A} = 2 \cdot \frac{1}{\cos A} \cdot \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2\operatorname{cosec} A = \text{RHS}$$

Note the two minus signs cancelling in the middle step. Losing one of them is the usual way this proof goes wrong, and the symptom is an answer of $-2\operatorname{cosec} A$.

Question 3

If $\tan A = \dfrac{3}{4}$, then $\sin A \cos A = \dfrac{12}{25}$

Solution. $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac34$, so take the opposite side as $3$ and the adjacent as $4$. By Pythagoras the hypotenuse is

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

the familiar $3$–$4$–$5$ triangle. Then

$$\sin A = \frac{3}{5}, \qquad \cos A = \frac{4}{5}, \qquad \sin A \cos A = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}$$

An identity-only route, for the practice: dividing $\sin A\cos A$ by $\sin^2 A + \cos^2 A$ (which is $1$, so this changes nothing) and then dividing top and bottom by $\cos^2 A$:

$$\sin A\cos A = \frac{\sin A\cos A}{\sin^2 A + \cos^2 A} = \frac{\tan A}{\tan^2 A + 1} = \frac{\frac34}{\frac{9}{16} + 1} = \frac{\frac34}{\frac{25}{16}} = \frac34 \cdot \frac{16}{25} = \frac{12}{25}$$

The same answer. The triangle is faster here, but the algebraic version is the one that survives when the ratio is a letter rather than a number.

Question 4

$$(\sin\alpha + \cos\alpha)(\tan\alpha + \cot\alpha) = \sec\alpha + \operatorname{cosec}\alpha$$

Solution. Deal with the second bracket first, since it is the one with the fractions:

$$\tan\alpha + \cot\alpha = \frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\sin\alpha} = \frac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha} = \frac{1}{\sin\alpha\cos\alpha}$$

That single simplification is the whole proof. The left side becomes

$$(\sin\alpha + \cos\alpha) \cdot \frac{1}{\sin\alpha\cos\alpha} = \frac{\sin\alpha}{\sin\alpha\cos\alpha} + \frac{\cos\alpha}{\sin\alpha\cos\alpha}$$

Cancel one factor in each term:

$$= \frac{1}{\cos\alpha} + \frac{1}{\sin\alpha} = \sec\alpha + \operatorname{cosec}\alpha = \text{RHS}$$

$\tan\alpha + \cot\alpha = \dfrac{1}{\sin\alpha\cos\alpha}$ is worth committing to memory outright. It appears again in Question 7 of this exercise and in Question 2 of Exercise 8.4.

Question 5

$$\left(\sqrt{3} + 1\right)(3 – \cot 30^\circ) = \tan^3 60^\circ – 2\sin 60^\circ$$

Solution. This one is pure arithmetic with the standard table — no identity is involved. Evaluate each side.

Left side. $\cot 30^\circ = \sqrt3$, so

$$\left(\sqrt3 + 1\right)\left(3 – \sqrt3\right)$$

Factor $\sqrt3$ out of the second bracket, since $3 = \sqrt3 \cdot \sqrt3$:

$$= \left(\sqrt3+1\right) \cdot \sqrt3\left(\sqrt3 – 1\right) = \sqrt3\left(\sqrt3+1\right)\left(\sqrt3-1\right) = \sqrt3\left(3 – 1\right) = 2\sqrt3$$

using the difference of squares $(\sqrt3+1)(\sqrt3-1) = 3 – 1 = 2$.

Right side. $\tan 60^\circ = \sqrt3$ and $\sin 60^\circ = \dfrac{\sqrt3}{2}$, so

$$\left(\sqrt3\right)^3 – 2 \cdot \frac{\sqrt3}{2} = 3\sqrt3 – \sqrt3 = 2\sqrt3$$

using $(\sqrt3)^3 = (\sqrt3)^2 \cdot \sqrt3 = 3\sqrt3$.

Both sides equal $2\sqrt3 \approx 3.464$.

Question 6

$$1 + \frac{\cot^2\alpha}{1 + \operatorname{cosec}\alpha} = \operatorname{cosec}\alpha$$

Solution. The key is to recognise the numerator as a difference of squares. From $\operatorname{cosec}^2\alpha = 1 + \cot^2\alpha$:

$$\cot^2\alpha = \operatorname{cosec}^2\alpha – 1 = (\operatorname{cosec}\alpha – 1)(\operatorname{cosec}\alpha + 1)$$

Substituting that into the fraction, the factor $(\operatorname{cosec}\alpha + 1)$ cancels against the denominator:

$$\frac{\cot^2\alpha}{1 + \operatorname{cosec}\alpha} = \frac{(\operatorname{cosec}\alpha – 1)(\operatorname{cosec}\alpha + 1)}{1 + \operatorname{cosec}\alpha} = \operatorname{cosec}\alpha – 1$$

So the left side is

$$1 + \left(\operatorname{cosec}\alpha – 1\right) = \operatorname{cosec}\alpha = \text{RHS}$$

The general pattern. Whenever a denominator reads $1 \pm \text{(some ratio)}$ and the numerator is the square of the associated ratio, factorise. $\cot^2$ pairs with $\operatorname{cosec}$, $\tan^2$ with $\sec$, and $\sin^2$ or $\cos^2$ with each other via $\sin^2 = (1-\cos)(1+\cos)$. Three identities, each generating a cancellation.

Question 7

$$\tan\theta + \tan(90^\circ – \theta) = \sec\theta \ \sec(90^\circ – \theta)$$

Solution. Convert both complementary terms to the same angle. Since $\tan(90^\circ – \theta) = \cot\theta$ and $\sec(90^\circ – \theta) = \operatorname{cosec}\theta$, the claim becomes

$$\tan\theta + \cot\theta = \sec\theta\operatorname{cosec}\theta$$

Now the left side, exactly as in Question 4:

$$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$$

And the right side:

$$\sec\theta\operatorname{cosec}\theta = \frac{1}{\cos\theta} \cdot \frac{1}{\sin\theta} = \frac{1}{\sin\theta\cos\theta}$$

The two agree.

So this is Question 4’s identity with a complementary-angle wrapper. Strip the wrapper first and the proof is one you have already written.

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

Question 8

Find the angle of elevation of the sun when the shadow of a pole $h$ metres high is $\sqrt{3}\,h$ metres long.

Solution. The pole stands vertically and its shadow lies flat on the ground, so the two meet at a right angle. Calling the elevation $\theta$, the pole is the side opposite $\theta$ and the shadow the side adjacent, so the tangent connects them:

$$\tan\theta = \frac{\text{height of pole}}{\text{length of shadow}} = \frac{h}{\sqrt3\,h} = \frac{1}{\sqrt3}$$

The $h$ cancels — the answer does not depend on how tall the pole is, only on the ratio of shadow to height, which is what makes this a well-posed question with a letter in it.

And $\tan\theta = \dfrac{1}{\sqrt3}$ gives $\theta = 30^\circ$.

The shadow being longer than the pole (by a factor of $\sqrt3 \approx 1.73$) tells you before any calculation that the Sun is low, so the elevation must be under $45^\circ$.

Question 9

If $\sqrt{3}\,\tan\theta = 1$, then find the value of $\sin^2\theta – \cos^2\theta$.

Solution. From the condition,

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

Then

$$\sin^2 30^\circ – \cos^2 30^\circ = \left(\frac12\right)^2 – \left(\frac{\sqrt3}{2}\right)^2 = \frac14 – \frac34 = -\frac12$$

The answer is negative, and that is not a slip. $\theta = 30^\circ$ is below $45^\circ$, where the cosine is the larger of the two ratios, so $\sin^2\theta – \cos^2\theta$ must come out below zero. Anyone who “corrects” it to $+\frac12$ has lost the question.

A route that never names the angle, useful when the tangent is not a table value:

$$\sin^2\theta – \cos^2\theta = \frac{\sin^2\theta – \cos^2\theta}{\sin^2\theta + \cos^2\theta} = \frac{\tan^2\theta – 1}{\tan^2\theta + 1} = \frac{\frac13 – 1}{\frac13 + 1} = \frac{-\frac23}{\frac43} = -\frac12$$

Dividing by $\sin^2\theta + \cos^2\theta$ is legitimate because that quantity is $1$; it is the standard way to turn a homogeneous expression into one in $\tan\theta$ alone.

Question 10

A ladder $15$ metres long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, find the height of the wall.

Solution. Read the angle carefully: it is measured with the wall, not with the ground. That single detail decides the whole question.

The ladder, the wall and the ground form a right-angled triangle, right angled where the wall meets the ground. The ladder is the hypotenuse, of length $15$ m. The $60^\circ$ angle sits at the top, between the ladder and the wall.

Relative to that $60^\circ$ angle, the wall is the adjacent side, so the ratio to use is the cosine:

$$\cos 60^\circ = \frac{\text{height of wall}}{\text{length of ladder}} = \frac{h}{15}$$

$$h = 15\cos 60^\circ = 15 \times \frac12 = \frac{15}{2} = 7.5 \text{ m}$$

The wrong answer this question is set to catch is $15\sin 60^\circ = \dfrac{15\sqrt3}{2} \approx 12.99$ m, which comes from assuming the $60^\circ$ is at the foot. Sketching the triangle and marking where the angle actually sits prevents it.

A check that the answer is sensible: the ladder leans far out from the wall (the angle at the foot is only $30^\circ$), so it cannot reach high, and $7.5$ m is half its length. If the angle at the top were smaller — a steeper ladder — the wall would be taller.

Question 11

Simplify $\left(1 + \tan^2\theta\right)(1 – \sin\theta)(1 + \sin\theta)$

Solution. Take the last two brackets together first — they are a difference of squares:

$$(1 – \sin\theta)(1 + \sin\theta) = 1 – \sin^2\theta = \cos^2\theta$$

And the first bracket is the Pythagorean identity in its $\sec$ form:

$$1 + \tan^2\theta = \sec^2\theta = \frac{1}{\cos^2\theta}$$

Multiplying:

$$\frac{1}{\cos^2\theta} \times \cos^2\theta = 1$$

The expression simplifies to $1$ for every angle — a constant, with no $\theta$ left in it.

The question rewards recognising two standard forms rather than expanding. Multiplying the three brackets out gives $1 + \tan^2\theta – \sin^2\theta – \sin^2\theta\tan^2\theta$, which does eventually collapse to $1$, but only after converting to sines and cosines and combining four terms.

Question 12

If $2\sin^2\theta – \cos^2\theta = 2$, then find the value of $\theta$.

Solution. Reduce to a single ratio. Using $\cos^2\theta = 1 – \sin^2\theta$:

$$2\sin^2\theta – \left(1 – \sin^2\theta\right) = 2$$

$$2\sin^2\theta – 1 + \sin^2\theta = 2 \implies 3\sin^2\theta = 3 \implies \sin^2\theta = 1$$

So $\sin\theta = 1$ (taking the positive root, since $\theta$ is acute), which gives

$$\theta = 90^\circ$$

Check it in the original: $2\sin^2 90^\circ – \cos^2 90^\circ = 2(1) – 0 = 2$. Correct.

This is an equation with a solution right at the edge of the permitted range, which is a good reason to verify rather than assume. Note also that $\sin^2\theta = 1$ forces the answer — there is no other acute angle whose sine squared is $1$.

Question 13

Show that $\dfrac{\cos^2(45^\circ + \theta) + \cos^2(45^\circ – \theta)}{\tan(60^\circ + \theta)\ \tan(30^\circ – \theta)} = 1$

Solution. Both the numerator and the denominator are complementary pairs. Check the sums:

$$(45^\circ + \theta) + (45^\circ – \theta) = 90^\circ, \qquad (60^\circ + \theta) + (30^\circ – \theta) = 90^\circ$$

The numerator. Using $\cos(90^\circ – x) = \sin x$ with $x = 45^\circ + \theta$:

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

So the numerator becomes

$$\cos^2(45^\circ + \theta) + \sin^2(45^\circ + \theta) = 1$$

The denominator. Using $\tan(90^\circ – x) = \cot x$ with $x = 60^\circ + \theta$:

$$\tan(30^\circ – \theta) = \tan\big[90^\circ – (60^\circ + \theta)\big] = \cot(60^\circ + \theta)$$

So the denominator becomes

$$\tan(60^\circ + \theta) \cdot \cot(60^\circ + \theta) = \tan(60^\circ + \theta) \cdot \frac{1}{\tan(60^\circ + \theta)} = 1$$

Therefore the whole expression is $\dfrac{1}{1} = 1$, for every $\theta$.

Both halves came out as $1$, which is the recurring design of this unit: a complementary pair either sums to $1$ through the Pythagorean identity or multiplies to $1$ through the reciprocal relation.

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

Question 14

An observer $1.5$ metres tall is $20.5$ metres away from a tower $22$ metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Solution. The angle is measured from the eye, not from the ground, so the observer’s height has to be subtracted before anything else. Drawing a horizontal line from the eye across to the tower gives a right-angled triangle whose:

• horizontal leg is the distance to the tower, $20.5$ m
• vertical leg is the part of the tower above eye level, which is $22 – 1.5 = 20.5$ m

Both legs are $20.5$ m. So

$$\tan\theta = \frac{20.5}{20.5} = 1 \implies \theta = 45^\circ$$

The numbers were chosen so that $22 – 1.5$ lands exactly on the given distance, which makes the triangle isosceles and the answer a table value. If your two legs are not coming out equal, the likeliest cause is using $22$ instead of $20.5$ for the vertical — that would give $\tan\theta = \frac{22}{20.5} \approx 1.073$ and an angle of about $47^\circ$, close enough to $45^\circ$ to look plausible and wrong.

Subtracting the observer’s height is the standard first step in every height-and-distance problem where an eye level is mentioned. Where the question says nothing about the observer’s height, the angle is taken from the ground and no subtraction is needed.

Question 15

Show that $\tan^4\theta + \tan^2\theta = \sec^4\theta – \sec^2\theta$

Solution. Work on the left side and factorise:

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

The bracket is the Pythagorean identity: $\tan^2\theta + 1 = \sec^2\theta$. And the factor in front can be written the same way round: $\tan^2\theta = \sec^2\theta – 1$. Substituting both:

$$= \left(\sec^2\theta – 1\right)\sec^2\theta = \sec^4\theta – \sec^2\theta = \text{RHS}$$

Two applications of one identity, in opposite directions — that is the whole proof, and it is worth noticing that the identity was used to introduce a $\sec^2\theta$ in one place and to eliminate a $\tan^2\theta$ in another. Identities run both ways, and choosing the direction is the skill.

The structure is also visible from the other end. The right side factorises as $\sec^2\theta(\sec^2\theta – 1) = \sec^2\theta\tan^2\theta$, and the left side factorises as $\tan^2\theta(\tan^2\theta+1) = \tan^2\theta\sec^2\theta$. Both meet at the same middle expression, $\tan^2\theta\sec^2\theta$, which is the cleanest way to present the proof if you are asked to show it rather than derive it.

Common mistakes

• Question 10 — assuming the angle is at the foot of the ladder. The question says the ladder makes $60^\circ$ with the wall, so the wall is the side adjacent to that angle and the height is $15\cos 60^\circ = 7.5$ m. Reading it as $60^\circ$ with the ground gives $12.99$ m, a plausible-looking wrong answer. Always mark the angle on a sketch before choosing a ratio.
• Question 14 — measuring the elevation from the ground. The angle is from the observer’s eye, so the vertical leg is $22 – 1.5 = 20.5$ m, not $22$ m. The error shifts the answer from $45^\circ$ to about $47^\circ$, close enough to slip through unnoticed.
• Question 9 — changing the sign of a negative answer. $\sin^2 30^\circ – \cos^2 30^\circ = -\frac12$, and it must be negative because $30^\circ$ lies below the crossing point at $45^\circ$ where cosine is the larger ratio. A negative answer to a trigonometric question is not automatically a mistake.
• Questions 1, 2 and 4 — manipulating both sides at once. In a proof you may not assume what you are proving. Fix one side, transform it, and arrive at the other; if you must work on both, transform each independently to a common third expression and say so.
• Questions 6 and 15 — expanding instead of factorising. $\cot^2\alpha = (\operatorname{cosec}\alpha – 1)(\operatorname{cosec}\alpha + 1)$ and $\tan^2\theta + 1 = \sec^2\theta$ turn both questions into one cancellation each. Multiplying everything out gets there too, but through four times the algebra.
• Question 11 — expanding three brackets. Two of them form a difference of squares, giving $\cos^2\theta$, and the third is $\sec^2\theta$. Look for the standard forms before reaching for the distributive law.
• Question 2 — losing one of the two minus signs. The numerator is $-2\sec A$ and the denominator is $-\tan^2 A$; both are negative and they cancel. An answer of $-2\operatorname{cosec} A$ means exactly one of them was dropped.

Practise next

• Exercise 8.4 — eighteen long-answer questions, where these identities reappear in harder forms and the height-and-distance problems grow to two observation points.
• Exercise 8.2 — worth revisiting after Questions 8, 10 and 14 here, since its Questions 7 to 12 are the same elevation reasoning stated as claims to be judged rather than problems to be solved.
• Exercise 8.1 — the fastest revision of the specific-angle table, which Questions 5, 9, 10 and 12 here all depend on.