Introduction to Trigonometry

Introduction to Trigonometry

Introduction to Trigonometry carries 6 or 7 marks in the board paper: 6 in the 2026-27 sample paper and 7 in the 2025 and both 2026 board papers.

The marks usually come as one or two 1-mark MCQs (a ratio from a given ratio, values at standard angles, a quick simplification), a 2-mark question (evaluate an expression, or find $A$ and $B$), and a 3-mark identity proof, sometimes set as "find the error in this working". There has been no 5-mark question from this chapter, so the long questions here are case-based and 4-mark items.

Where marks are usually lost:

• taking the wrong side as the base or the perpendicular for the angle asked;
• using $\sin^2\theta + \cos^2\theta = 1$ when the two angles are different, or writing $\sin(A + B) = \sin A + \sin B$;
• in proofs, cross-multiplying or moving terms across the = sign, or not showing which identity was used;
• stopping at $3\theta = 45^\circ$ instead of finding $\theta$.

Revise in 5 Minutes

Ratios of an acute angle $A$ in a right triangle

• $\text{cosec}\,A = \frac{1}{\sin A}$, $\sec A = \frac{1}{\cos A}$, $\cot A = \frac{1}{\tan A}$
• $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$

Values to know by heart

Identities

• $\sin^2 A + \cos^2 A = 1$
• $\sec^2 A - \tan^2 A = 1$, so $(\sec A - \tan A)(\sec A + \tan A) = 1$
• $\text{cosec}^2 A - \cot^2 A = 1$, so $(\text{cosec}\,A - \cot A)(\text{cosec}\,A + \cot A) = 1$

Quick facts for an acute angle: from $0^\circ$ to $90^\circ$, $\sin A$ increases and $\cos A$ decreases; both lie between 0 and 1; $\sec A$ and $\text{cosec}\,A$ are more than 1; $\tan A$ can be any positive number.

One ratio given? Draw the right triangle, find the third side by Pythagoras, then read off any ratio.

Proving an identity

1. Start from the more complicated side. Don't cross-multiply or move terms across =; reducing both sides separately to the same expression is fine.
2. If stuck, change everything into $\sin$ and $\cos$.
3. Useful algebra: $a^2 + b^2 = (a + b)^2 - 2ab$ and $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with $a + b = \sin^2 A + \cos^2 A = 1$.
4. End with "= RHS".

Traps

• $\sin(A + B) \neq \sin A + \sin B$; check with $A = B = 30^\circ$.
• $\sin^2 60^\circ + \cos^2 30^\circ$ is not 1: the angles differ.
• If $\cos 3\theta = \frac{1}{\sqrt{2}}$, then $3\theta = 45^\circ$ and $\theta = 15^\circ$.
• $\tan A \times \cot A = 1$, not $\tan A + \cot A$.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

Evaluate: $\frac{4\cot^2 60^\circ + \sec^2 30^\circ - 2\sin^2 45^\circ}{\sin^2 60^\circ + \cos^2 45^\circ}$

Answer.

1. Numerator $= 4 \times \frac{1}{3} + \frac{4}{3} - 2 \times \frac{1}{2} = \frac{5}{3}$ — 1 mark
2. Denominator $= \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$; value $= \frac{5}{3} \div \frac{5}{4} = \frac{4}{3}$ — 1 mark

Question 2 (2 marks)

Find $x$ and $y$ if $x\tan 45^\circ + y\cos 60^\circ = 7$ and $x\sin 30^\circ - y\sec 60^\circ = -1$.

Answer.

1. Putting in the values: $x + \frac{y}{2} = 7$ and $\frac{x}{2} - 2y = -1$, that is, $2x + y = 14$ and $x - 4y = -2$ — 1 mark
2. From the first, $y = 14 - 2x$; then $x - 56 + 8x = -2$, so $x = 6$ and $y = 2$ — 1 mark

Question 3 (2 marks)

Neha was asked to find the value of $\sin^2 60^\circ + \cos^2 30^\circ$. She wrote: “By the identity $\sin^2\theta + \cos^2\theta = 1$, the value is 1.” Is she right? Find the correct value.

Answer.

Model answer:

No, Neha is not right.

The identity $\sin^2\theta + \cos^2\theta = 1$ is true only when both ratios are of the same angle $\theta$. Here one ratio is of $60^\circ$ and the other is of $30^\circ$, so the identity cannot be used.

Putting in the values: $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. So $\sin^2 60^\circ + \cos^2 30^\circ = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}$.

Marking scheme:

1. No; the identity holds only when both ratios are of the same angle, and here the angles are $60^\circ$ and $30^\circ$ — 1 mark
2. $\sin^2 60^\circ + \cos^2 30^\circ = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}$ — 1 mark

Question 4 (2 marks)

If $5\sin\theta = 3$, where $\theta$ is acute, find the value of $\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta}$.

Answer.

1. $\sin\theta = \frac{3}{5}$, base $= \sqrt{25 - 9} = 4$; so $\sec\theta = \frac{5}{4}$ and $\tan\theta = \frac{3}{4}$ — 1 mark
2. Value $= \frac{\frac{5}{4} - \frac{3}{4}}{\frac{5}{4} + \frac{3}{4}} = \frac{2}{8} = \frac{1}{4}$ — 1 mark

Question 5 (2 marks)

Prove that $\tan^2 A - \sin^2 A = \tan^2 A \sin^2 A$.

Answer.

Model answer:

LHS $= \tan^2 A - \sin^2 A = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A$.

Taking $\sin^2 A$ common and using the common denominator $\cos^2 A$: LHS $= \frac{\sin^2 A (1 - \cos^2 A)}{\cos^2 A}$.

Since $1 - \cos^2 A = \sin^2 A$, LHS $= \frac{\sin^2 A}{\cos^2 A} \times \sin^2 A = \tan^2 A \sin^2 A =$ RHS.

Hence proved.

Marking scheme:

1. LHS $= \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \frac{\sin^2 A (1 - \cos^2 A)}{\cos^2 A}$ — 1 mark
2. $= \frac{\sin^2 A}{\cos^2 A} \times \sin^2 A = \tan^2 A \sin^2 A =$ RHS — 1 mark

Question 6 (3 marks)

In $\triangle ABC$, right-angled at $B$, $\tan A = \frac{5}{12}$ and the perimeter of the triangle is 60 cm. Find the lengths of its sides and the value of $\sin C + \cos C$.

Answer.

1. $\tan A = \frac{BC}{AB} = \frac{5}{12}$: let $BC = 5k$, $AB = 12k$; then $AC = \sqrt{25k^2 + 144k^2} = 13k$ — 1 mark
2. $5k + 12k + 13k = 60$ gives $k = 2$: $BC = 10$ cm, $AB = 24$ cm, $AC = 26$ cm — 1 mark
3. $\sin C = \frac{AB}{AC} = \frac{12}{13}$, $\cos C = \frac{BC}{AC} = \frac{5}{13}$, so $\sin C + \cos C = \frac{17}{13}$ — 1 mark

Question 7 (3 marks)

Prove that $\frac{1 + \cos A + \sin A}{1 + \cos A - \sin A} = \frac{1 + \sin A}{\cos A}$.

Answer.

Model answer:

We work on the LHS only. Group it as $\frac{(1 + \cos A) + \sin A}{(1 + \cos A) - \sin A}$ and multiply the numerator and the denominator by $(1 + \cos A) + \sin A$.

Denominator: $(1 + \cos A)^2 - \sin^2 A = 1 + 2\cos A + \cos^2 A - \sin^2 A$. Using $1 - \sin^2 A = \cos^2 A$, this is $2\cos^2 A + 2\cos A = 2\cos A(1 + \cos A)$.

Numerator: $(1 + \cos A + \sin A)^2 = 1 + \cos^2 A + \sin^2 A + 2\cos A + 2\sin A + 2\sin A\cos A$. Using $\sin^2 A + \cos^2 A = 1$, this is $2 + 2\cos A + 2\sin A + 2\sin A\cos A = 2(1 + \cos A) + 2\sin A(1 + \cos A) = 2(1 + \cos A)(1 + \sin A)$.

So LHS $= \frac{2(1 + \cos A)(1 + \sin A)}{2\cos A(1 + \cos A)} = \frac{1 + \sin A}{\cos A} =$ RHS. Hence proved.

Marking scheme:

1. Multiply the numerator and the denominator of the LHS by $(1 + \cos A) + \sin A$; denominator $= (1 + \cos A)^2 - \sin^2 A = 1 + 2\cos A + \cos^2 A - \sin^2 A = 2\cos A(1 + \cos A)$ — 1 mark
2. Numerator $= (1 + \cos A + \sin A)^2 = 2 + 2\cos A + 2\sin A + 2\sin A\cos A = 2(1 + \cos A)(1 + \sin A)$ — 1 mark
3. LHS $= \frac{2(1 + \cos A)(1 + \sin A)}{2\cos A(1 + \cos A)} = \frac{1 + \sin A}{\cos A} =$ RHS — 1 mark

Question 8 (3 marks)

Prove that $2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 = 0$.

Answer.

Model answer:

Let $a = \sin^2\theta$ and $b = \cos^2\theta$, so that $a + b = 1$.

Using $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$: $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$.

Using $a^2 + b^2 = (a + b)^2 - 2ab$: $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$.

LHS $= 2(1 - 3\sin^2\theta\cos^2\theta) - 3(1 - 2\sin^2\theta\cos^2\theta) + 1 = 2 - 6\sin^2\theta\cos^2\theta - 3 + 6\sin^2\theta\cos^2\theta + 1 = 0 =$ RHS.

Hence proved.

Marking scheme:

1. $\sin^6\theta + \cos^6\theta = (\sin^2\theta + \cos^2\theta)^3 - 3\sin^2\theta\cos^2\theta(\sin^2\theta + \cos^2\theta) = 1 - 3\sin^2\theta\cos^2\theta$ — 1 mark
2. $\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - 2\sin^2\theta\cos^2\theta$ — 1 mark
3. LHS $= 2 - 6\sin^2\theta\cos^2\theta - 3 + 6\sin^2\theta\cos^2\theta + 1 = 0 =$ RHS — 1 mark

Question 9 (3 marks)

Prove that $\frac{\cos A}{1 + \sin A} + \tan A = \sec A$.

Answer.

Model answer:

LHS $= \frac{\cos A}{1 + \sin A} + \frac{\sin A}{\cos A}$. Taking the common denominator $(1 + \sin A)\cos A$:

LHS $= \frac{\cos^2 A + \sin A + \sin^2 A}{(1 + \sin A)\cos A}$.

Since $\sin^2 A + \cos^2 A = 1$, the numerator is $1 + \sin A$. So LHS $= \frac{1 + \sin A}{(1 + \sin A)\cos A} = \frac{1}{\cos A} = \sec A =$ RHS.

Hence proved.

Marking scheme:

1. LHS $= \frac{\cos A}{1 + \sin A} + \frac{\sin A}{\cos A} = \frac{\cos^2 A + \sin A(1 + \sin A)}{(1 + \sin A)\cos A}$ — 1 mark
2. Numerator $= \cos^2 A + \sin^2 A + \sin A = 1 + \sin A$ — 1 mark
3. LHS $= \frac{1 + \sin A}{(1 + \sin A)\cos A} = \frac{1}{\cos A} = \sec A =$ RHS — 1 mark

Question 10 (3 marks)

If $\sin\theta - \cos\theta = \frac{1}{2}$, where $\theta$ is acute, find the values of (i) $\sin\theta\cos\theta$, (ii) $\sin\theta + \cos\theta$ and (iii) $\tan\theta + \cot\theta$.

Answer.

1. (i) Squaring, $1 - 2\sin\theta\cos\theta = \frac{1}{4}$, so $\sin\theta\cos\theta = \frac{3}{8}$ — 1 mark
2. (ii) $(\sin\theta + \cos\theta)^2 = 1 + 2 \times \frac{3}{8} = \frac{7}{4}$; both are positive, so $\sin\theta + \cos\theta = \frac{\sqrt{7}}{2}$ — 1 mark
3. (iii) $\tan\theta + \cot\theta = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = 1 \div \frac{3}{8} = \frac{8}{3}$ — 1 mark

Long Answer and Case-Based Questions

Question 11 (4 marks)

A furniture shop in Pune uses a loading ramp to roll heavy cartons into its delivery van. The figure shows the side view. The ramp $AC$ rises $BC = 0.7$ m (the height of the van floor) over a horizontal distance $AB = 2.4$ m, and it makes an angle $\theta$ with the ground.

(i) Find the length of the ramp $AC$. (1 mark)

Answer.

1. $AC = \sqrt{2.4^2 + 0.7^2} = \sqrt{5.76 + 0.49} = \sqrt{6.25} = 2.5$ m — 1 mark

(ii) Find $\sin\theta$ and $\cos\theta$. (1 mark)

Answer.

1. $\sin\theta = \frac{0.7}{2.5} = \frac{7}{25}$ and $\cos\theta = \frac{2.4}{2.5} = \frac{24}{25}$ — 1 mark

(iii) Find the value of $\sec\theta + \tan\theta$, and check that $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$. (2 marks)

Answer.

1. $\sec\theta = \frac{25}{24}$, $\tan\theta = \frac{7}{24}$; so $\sec\theta + \tan\theta = \frac{32}{24} = \frac{4}{3}$ — 1 mark
2. $\sec\theta - \tan\theta = \frac{18}{24} = \frac{3}{4}$ and $\frac{4}{3} \times \frac{3}{4} = 1$; verified — 1 mark

OR

(iii) The same 2.5 m ramp is later used to load a truck whose floor is 1.5 m above the ground. If the ramp now makes an angle $\phi$ with the ground, find $\tan\phi$ and say which slope is steeper. (2 marks)

Answer.

1. Horizontal distance $= \sqrt{2.5^2 - 1.5^2} = \sqrt{4} = 2$ m, so $\tan\phi = \frac{1.5}{2} = \frac{3}{4}$ — 1 mark
2. $\tan\theta = \frac{7}{24}$ is less than $\frac{3}{4}$, so on the truck the ramp rises more for each metre across: that slope is steeper — 1 mark

Question 12 (4 marks)

Tanvi's geometry box has two set squares, shown in the figure. In set square $ABC$, $\angle B = 90^\circ$, $\angle A = 30^\circ$ and the longest side $AC = 12$ cm. In set square $PQR$, $\angle Q = 90^\circ$, $\angle P = 45^\circ$ and the longest side $PR = 10$ cm.

(i) Find the length of $BC$. (1 mark)

Answer.

1. $\sin 30^\circ = \frac{BC}{AC}$, so $BC = 12 \times \frac{1}{2} = 6$ cm — 1 mark

(ii) Find the length of $AB$. (1 mark)

Answer.

1. $\cos 30^\circ = \frac{AB}{AC}$, so $AB = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$ cm — 1 mark

(iii) Find the lengths of $PQ$ and $QR$, and the area of the set square $PQR$. (2 marks)

Answer.

1. $PQ = PR\cos 45^\circ = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}$ cm and $QR = PR\sin 45^\circ = 5\sqrt{2}$ cm — 1 mark
2. Area $= \frac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} = 25$ cm$^2$ — 1 mark

OR

(iii) Tanvi draws the perpendicular $BD$ from $B$ to $AC$ on the set square $ABC$. Using $AB$ from part (ii), find $BD$ and $AD$. (2 marks)

Answer.

1. In right $\triangle ADB$, $\angle A = 30^\circ$: $BD = AB\sin 30^\circ = 6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3}$ cm — 1 mark
2. $AD = AB\cos 30^\circ = 6\sqrt{3} \times \frac{\sqrt{3}}{2} = 9$ cm — 1 mark

Question 13 (4 marks)

For an acute angle $\theta$, it is given that $\sec\theta - \tan\theta = \frac{1}{9}$. Answer the following.

(i) Find the value of $\sec\theta + \tan\theta$. (1 mark)

Answer.

1. $\sec^2\theta - \tan^2\theta = 1$ gives $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1$, so $\sec\theta + \tan\theta = 9$ — 1 mark

(ii) Find $\sec\theta$ and $\tan\theta$. (1 mark)

Answer.

1. Adding: $2\sec\theta = 9 + \frac{1}{9} = \frac{82}{9}$, so $\sec\theta = \frac{41}{9}$; subtracting: $\tan\theta = \frac{40}{9}$ — 1 mark

(iii) Find $\sin\theta$ and $\cos\theta$, and verify that $\sin^2\theta + \cos^2\theta = 1$. (2 marks)

Answer.

1. $\cos\theta = \frac{1}{\sec\theta} = \frac{9}{41}$ and $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{40}{9} \times \frac{9}{41} = \frac{40}{41}$ — 1 mark
2. $\frac{1600}{1681} + \frac{81}{1681} = \frac{1681}{1681} = 1$; verified — 1 mark

Question 14 (4 marks)

This question is about the expression $\sqrt{\sec^2\theta + \text{cosec}^2\theta}$ for an acute angle $\theta$.

(i) Prove that $\sqrt{\sec^2\theta + \text{cosec}^2\theta} = \tan\theta + \cot\theta$, where $\theta$ is an acute angle. (2 marks)

Answer.

Model answer:

LHS: $\sec^2\theta + \text{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$.

For an acute angle, $\sin\theta$ and $\cos\theta$ are positive, so LHS $= \sqrt{\frac{1}{\sin^2\theta\cos^2\theta}} = \frac{1}{\sin\theta\cos\theta}$.

RHS $= \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$.

LHS = RHS. Hence proved.

Marking scheme:

1. $\sec^2\theta + \text{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$ — 1 mark
2. So LHS $= \frac{1}{\sin\theta\cos\theta}$; RHS $= \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$ — 1 mark

(ii) Verify the result of (i) for $\theta = 30^\circ$. (1 mark)

Answer.

1. LHS $= \sqrt{\frac{4}{3} + 4} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$; RHS $= \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{4}{\sqrt{3}}$; equal — 1 mark

(iii) If $\tan\theta + \cot\theta = 2$, find the acute angle $\theta$. (1 mark)

Answer.

1. $\tan\theta + \frac{1}{\tan\theta} = 2$ gives $(\tan\theta - 1)^2 = 0$, so $\tan\theta = 1$ and $\theta = 45^\circ$ — 1 mark