Introduction to Trigonometry · Class 10 Mathematics: 173 MCQs with answers

All 173 multiple-choice questions from this kit, free and with no sign-up.

Warm-up · 51 questions

Q1. In a right triangle, the hypotenuse is the side

Beginner
A facing the chosen acute angle B touching the chosen angle C facing the right angle D that is shortest
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facing the right angle

It is always the longest side, and it never changes with the angle.

Q2. $\sin A$ is defined as

Beginner
A $\frac{\text{adjacent}}{\text{hypotenuse}}$ B $\frac{\text{opposite}}{\text{hypotenuse}}$ C $\frac{\text{opposite}}{\text{adjacent}}$ D $\frac{\text{hypotenuse}}{\text{opposite}}$
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$\frac{\text{opposite}}{\text{hypotenuse}}$

SOH — Sine, Opposite, Hypotenuse.

Q3. $\cos A$ is defined as

Beginner
A $\frac{\text{opposite}}{\text{hypotenuse}}$ B $\frac{\text{adjacent}}{\text{opposite}}$ C $\frac{\text{hypotenuse}}{\text{adjacent}}$ D $\frac{\text{adjacent}}{\text{hypotenuse}}$
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$\frac{\text{adjacent}}{\text{hypotenuse}}$

CAH — Cosine, Adjacent, Hypotenuse.

Q4. $\tan A$ is defined as

Beginner
A $\frac{\text{opposite}}{\text{hypotenuse}}$ B $\frac{\text{adjacent}}{\text{opposite}}$ C $\frac{\text{adjacent}}{\text{hypotenuse}}$ D $\frac{\text{opposite}}{\text{adjacent}}$
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$\frac{\text{opposite}}{\text{adjacent}}$

TOA — Tangent, Opposite, Adjacent.

Q5. $\operatorname{cosec} A$ is the reciprocal of

Beginner
A $\cos A$ B $\sin A$ C $\tan A$ D $\cot A$
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$\sin A$

Match the third letters: coSecant with Sine.

Q6. $\sec A$ is the reciprocal of

Beginner
A $\sin A$ B $\cos A$ C $\tan A$ D $\operatorname{cosec} A$
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$\cos A$

seCant with Cosine, despite how the names look.

Q7. $\cot A$ is the reciprocal of

Beginner
A $\tan A$ B $\sin A$ C $\cos A$ D $\sec A$
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$\tan A$

The one pairing that is not crossed.

Q8. In a triangle with opposite 3, adjacent 4 and hypotenuse 5, $\sin A$ is

Beginner
A $\frac45$ B $\frac34$ C $\frac35$ D $\frac53$
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$\frac35$

Opposite over hypotenuse.

Q9. In the same triangle, $\cos A$ is

Beginner
A $\frac35$ B $\frac43$ C $\frac54$ D $\frac45$
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$\frac45$

Adjacent over hypotenuse.

Q10. In the same triangle, $\tan A$ is

Beginner
A $\frac43$ B $\frac35$ C $\frac34$ D $\frac45$
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$\frac34$

Opposite over adjacent.

Q11. $\tan A$ can also be written as

Beginner
A $\frac{\cos A}{\sin A}$ B $\frac{\sin A}{\cos A}$ C $\sin A \cos A$ D $\sin A + \cos A$
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$\frac{\sin A}{\cos A}$

The common hypotenuse cancels from the two ratios.

Q12. Assertion (A): Two right triangles with the same acute angle give the same value of $\sin$ for that angle. Reason (R): The two triangles have the same area.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is true but R is false

Nothing fixes their areas as equal; what matters is that they are similar. The assertion holds, the reason is false.

Q13. Which side is never affected by which acute angle you choose?

Beginner
A the opposite side B the adjacent side C all three change D the hypotenuse
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the hypotenuse

It always faces the right angle.

Q14. In a triangle with opposite 5, adjacent 12 and hypotenuse 13, $\sin A$ is

Beginner
A $\frac{12}{13}$ B $\frac{5}{13}$ C $\frac{5}{12}$ D $\frac{13}{5}$
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$\frac{5}{13}$

Opposite over hypotenuse.

Q15. SOH-CAH-TOA is a way to remember

Beginner
A the three reciprocals B Pythagoras' theorem C the angle sum D the three main ratios
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the three main ratios

Sine-Opposite-Hypotenuse and so on.

Q16. In a right triangle, define $\sin A$, $\cos A$ and $\tan A$ in terms of the sides.

Beginner
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Q17. State the three reciprocal ratios, saying which ratio each one inverts.

Beginner
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Q18. $\sin 30°$ equals

Beginner
A $\frac12$ B $\frac{\sqrt3}{2}$ C $\frac{1}{\sqrt2}$ D 1
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$\frac12$

The smallest non-zero entry in the sine row.

Q19. $\cos 60°$ equals

Beginner
A $\frac{\sqrt3}{2}$ B $\frac{1}{\sqrt2}$ C 0 D $\frac12$
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$\frac12$

Cosine decreases, and at 60° it has reached a half.

Q20. $\sin 45°$ equals

Beginner
A $\frac12$ B $\frac{1}{\sqrt2}$ C $\frac{\sqrt3}{2}$ D 1
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$\frac{1}{\sqrt2}$

Equal legs give equal sine and cosine.

Q21. $\tan 45°$ equals

Beginner
A 0 B 1 C $\sqrt3$ D undefined
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1

The two legs are equal, so their ratio is 1.

Q22. $\cos 0°$ equals

Beginner
A 0 B $\frac12$ C 1 D undefined
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1

At 0° the adjacent side equals the hypotenuse.

Q23. $\sin 0°$ equals

Beginner
A 1 B $\frac12$ C 0 D undefined
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0

The opposite side has shrunk to nothing.

Q24. $\sin 90°$ equals

Beginner
A 0 B $\frac{\sqrt3}{2}$ C undefined D 1
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1

The opposite side has become the hypotenuse.

Q25. $\cos 90°$ equals

Beginner
A 1 B $\frac12$ C undefined D 0
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0

The adjacent side has vanished.

Q26. $\tan 90°$ is

Beginner
A infinity B undefined C 0 D 1
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undefined

It would require dividing by an adjacent side of length zero, which has no value.

Q27. $\tan 60°$ equals

Beginner
A $\frac{1}{\sqrt3}$ B 1 C $\sqrt3$ D $\frac{\sqrt3}{2}$
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$\sqrt3$

$\frac{\sqrt3/2}{1/2} = \sqrt3$.

Q28. $\tan 30°$ equals

Beginner
A $\sqrt3$ B 1 C $\frac12$ D $\frac{1}{\sqrt3}$
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$\frac{1}{\sqrt3}$

$\frac{1/2}{\sqrt3/2} = \frac1{\sqrt3}$.

Q29. Assertion (A): $\sin 45° = \cos 45°$. Reason (R): The 45° values come from a right triangle whose two legs are equal.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Equal legs make the opposite and adjacent sides interchangeable.

Q30. $\cos 30°$ equals

Beginner
A $\frac12$ B $\frac{1}{\sqrt2}$ C $\frac{\sqrt3}{2}$ D 1
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$\frac{\sqrt3}{2}$

Cosine starts at 1 and has only fallen a little by 30°.

Q31. $\sin 45°$ and $\cos 45°$ are

Beginner
A equal to each other B reciprocals C both 1 D both 0
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equal to each other

The two legs of the triangle are equal.

Q32. $\sec^2 45°$ equals

Beginner
A 1 B $\sqrt2$ C 2 D $\frac12$
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2

$\sec 45° = \sqrt2$, and its square is 2.

Q33. Assertion (A): The 30° and 60° values can both be read from one triangle. Reason (R): Bisecting an equilateral triangle produces a right triangle containing both a 30° and a 60° angle.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

One construction supplies four of the tabulated values at once.

Q34. Write the values of $\sin$, $\cos$ and $\tan$ for 30°, 45° and 60°.

Beginner
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Q35. Find the acute angle $\theta$ for which $\tan\theta = \sqrt3$.

Beginner
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Q36. $\sin^2\theta + \cos^2\theta$ equals

Beginner
A 1 B 0 C 2 D $\tan\theta$
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1

The first and most important identity.

Q37. $\sec^2\theta - \tan^2\theta$ equals

Beginner
A 0 B $-1$ C 1 D $\sec\theta$
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1

Identity 2, obtained by dividing identity 1 by $\cos^2\theta$.

Q38. $\operatorname{cosec}^2\theta - \cot^2\theta$ equals

Beginner
A 1 B 0 C $-1$ D $\cot\theta$
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1

Identity 3, from dividing by $\sin^2\theta$.

Q39. Identity 1 is proved from

Beginner
A the section formula B Pythagoras' theorem C the angle sum property D similar triangles alone
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Pythagoras' theorem

Divide $p^2 + b^2 = h^2$ by $h^2$.

Q40. To obtain $\sec^2\theta - \tan^2\theta = 1$, divide identity 1 by

Beginner
A $\sin^2\theta$ B $\cos^2\theta$ C $\tan^2\theta$ D 2
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$\cos^2\theta$

Cosine gives secant.

Q41. To obtain $\operatorname{cosec}^2\theta - \cot^2\theta = 1$, divide identity 1 by

Beginner
A $\cos^2\theta$ B $\cot^2\theta$ C $\sin^2\theta$ D 2
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$\sin^2\theta$

Sine gives cosecant.

Q42. $1 - \cos^2\theta$ equals

Beginner
A $\cos^2\theta$ B $\tan^2\theta$ C 1 D $\sin^2\theta$
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$\sin^2\theta$

A rearrangement of identity 1.

Q43. $1 + \tan^2\theta$ equals

Beginner
A $\operatorname{cosec}^2\theta$ B $\cot^2\theta$ C 1 D $\sec^2\theta$
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$\sec^2\theta$

A rearrangement of identity 2.

Q44. $1 + \cot^2\theta$ equals

Beginner
A $\sec^2\theta$ B $\operatorname{cosec}^2\theta$ C $\tan^2\theta$ D 1
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$\operatorname{cosec}^2\theta$

A rearrangement of identity 3.

Q45. An identity is a statement that is true

Beginner
A for one angle only B for every angle C for no angle D only for acute angles
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for every angle

That is what distinguishes it from an equation.

Q46. When proving an identity you should work on

Beginner
A both sides at once B neither side C the answer first D one side only
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one side only

Transforming both sides assumes the result.

Q47. Assertion (A): $\sin^2\theta + \cos^2\theta = 1$ is an identity. Reason (R): It is true for every value of $\theta$, not merely for particular ones.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Holding universally is exactly what makes a statement an identity.

Q48. $\sec^2\theta - 1$ equals

Beginner
A $\cot^2\theta$ B $\sin^2\theta$ C 0 D $\tan^2\theta$
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$\tan^2\theta$

Rearranging identity 2.

Q49. $\operatorname{cosec}^2\theta - 1$ equals

Beginner
A $\cot^2\theta$ B $\tan^2\theta$ C $\cos^2\theta$ D 0
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$\cot^2\theta$

Rearranging identity 3.

Q50. $\frac{\tan\theta}{\sin\theta}$ equals

Beginner
A $\cos\theta$ B $\operatorname{cosec}\theta$ C $\sec\theta$ D 1
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$\sec\theta$

$\frac{\sin\theta}{\cos\theta} \div \sin\theta = \frac1{\cos\theta}$.

Q51. State the three trigonometric identities.

Beginner
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Building up · 71 questions

Q52. Which of these is impossible?

Intermediate
A $\sin A = \frac54$ B $\sin A = \frac35$ C $\tan A = 5$ D $\sec A = 3$
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$\sin A = \frac54$

The hypotenuse is the longest side, so $\sin A$ cannot exceed 1.

Q53. The smallest possible value of $\sec A$ for an acute angle is

Intermediate
A 0 B $-1$ C there is none D 1
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1

It is the reciprocal of $\cos A$, which is at most 1.

Q54. Which pair of ratios has NO upper limit?

Intermediate
A $\tan A$ and $\cot A$ B $\sin A$ and $\cos A$ C $\sec A$ and $\operatorname{cosec} A$ D all are bounded
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$\tan A$ and $\cot A$

Neither involves the hypotenuse, so either side may be much longer.

Q55. $\sin^2 A$ means

Intermediate
A $\sin(A^2)$ B $(\sin A)^2$ C $\frac{1}{\sin A}$ D $2\sin A$
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$(\sin A)^2$

The power applies to the ratio, not to the angle.

Q56. $\sin^{-1} A$ means

Intermediate
A $\frac{1}{\sin A}$ B the angle whose sine is A C $\operatorname{cosec} A$ D $-\sin A$
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the angle whose sine is A

It is the inverse function; the reciprocal is cosec.

Q57. Taking the hypotenuse as 5 when $\sin A = \frac35$, the adjacent side is

Intermediate
A 4 B 3 C $\sqrt{34}$ D 2
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4

$\sqrt{25 - 9} = 4$.

Q58. A right triangle has $\sin A = \frac35$. Its cosine is

Intermediate
A $\frac53$ B $\frac34$ C $\frac45$ D $\frac54$
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$\frac45$

The adjacent side is 4, so $\cos A = \frac45$.

Q59. With $\sin A = \frac35$, the value of $\cot A$ is

Intermediate
A $\frac34$ B $\frac53$ C $\frac43$ D $\frac45$
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$\frac43$

$\tan A = \frac34$, and cot is its reciprocal.

Q60. If $\cos A = \frac{12}{13}$, then $\sin A$ is

Intermediate
A $\frac{13}{12}$ B $\frac{12}{5}$ C $\frac{5}{13}$ D $\frac{5}{12}$
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$\frac{5}{13}$

The opposite side is $\sqrt{169 - 144} = 5$.

Q61. If $\tan A = \frac12$, the hypotenuse of the simplest such triangle is

Intermediate
A $\sqrt5$ B 3 C $\sqrt3$ D 5
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$\sqrt5$

Opposite 1 and adjacent 2 give $\sqrt{1 + 4}$.

Q62. Why may you take the sides as 1 and 2 when $\tan A = \frac12$?

Intermediate
A the sides must be whole numbers B the hypotenuse must be $\sqrt5$ C the angle must be 30° D any such triangle is similar, so the ratios are the same
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any such triangle is similar, so the ratios are the same

Similar triangles share every ratio.

Q63. Assertion (A): $\sin^{-1} A$ is the same as $\operatorname{cosec} A$. Reason (R): $\operatorname{cosec} A$ is defined as $\frac{1}{\sin A}$.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

R is true, but A is false: $\sin^{-1}$ denotes the inverse function, not the reciprocal.

Q64. Assertion (A): When the acute angle is switched to the other one, the hypotenuse changes. Reason (R): 'Opposite' and 'adjacent' are named relative to the chosen angle.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

R is true, but the hypotenuse always faces the right angle and never changes — so A is false.

Q65. In that same triangle, $\sec A$ is

Intermediate
A $\frac{13}{5}$ B $\frac{13}{12}$ C $\frac{12}{13}$ D $\frac{5}{13}$
Show answer & explanation

$\frac{13}{12}$

$\sec A = \frac1{\cos A} = \frac{13}{12}$.

Q66. A triangle has $\sin A = \frac{7}{25}$. Its cosine is

Intermediate
A $\frac{25}{24}$ B $\frac{24}{25}$ C $\frac{7}{24}$ D $\frac{24}{7}$
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$\frac{24}{25}$

The adjacent side is $\sqrt{625 - 49} = 24$.

Q67. $\cot A$ can also be written as

Intermediate
A $\frac{\sin A}{\cos A}$ B $\frac{1}{\cos A}$ C $\frac{\cos A}{\sin A}$ D $\sin A \cos A$
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$\frac{\cos A}{\sin A}$

It is the reciprocal of $\tan A = \frac{\sin A}{\cos A}$.

Q68. Assertion (A): Given only $\cos A$, every other ratio of A can be found. Reason (R): The given ratio fixes two sides, and Pythagoras supplies the third.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Three sides in proportion determine all six ratios.

Q69. Explain why the value of $\sin A$ does not depend on the size of the right triangle.

Intermediate
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Q70. Given $\sin A = \frac{3}{5}$, find all the other five ratios.

Intermediate
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Q71. Given $\cos A = \frac{12}{13}$, find $\sin A$ and $\tan A$.

Intermediate
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Q72. Given $\tan A = \frac{1}{2}$, find $\sin A$ and $\cos A$.

Intermediate
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Q73. Explain why $\sin A$ can never be greater than 1.

Intermediate
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Q74. Explain why $\sec A$ can never be less than 1.

Intermediate
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Q75. Show that $\tan A \times \cot A = 1$ for any acute angle A.

Intermediate
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Q76. (a) In a right triangle ABC right-angled at B, explain how the three sides are named relative to angle A. (b) Define all six trigonometric ratios of A. (c) If $AB = 24$ cm, $BC = 7$ cm, find all six ratios of angle A.

Intermediate
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1½ marks — (a) all three sides named with the 'relative to the angle' point. 1½ marks — (b) all six definitions. 2 marks — (c) AC = 25 and all six values.

Q77. The 45° ratios are derived from

Intermediate
A a square cut along its diagonal B an equilateral triangle cut in half C a circle D a flattened triangle
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a square cut along its diagonal

That gives legs 1, 1 and hypotenuse $\sqrt2$.

Q78. The 30° and 60° ratios are derived from

Intermediate
A a square cut along its diagonal B a flattened triangle C an equilateral triangle cut in half D a semicircle
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an equilateral triangle cut in half

The perpendicular gives sides 1, $\sqrt3$ and 2.

Q79. Cutting an equilateral triangle of side 2 in half gives sides

Intermediate
A $1, \sqrt3, 2$ B $1, 2, 3$ C $2, 2, 2$ D $1, 1, \sqrt2$
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$1, \sqrt3, 2$

The base halves to 1 and the height is $\sqrt{4-1}$.

Q80. As the angle grows from 0° to 90°, $\cos$

Intermediate
A increases from 0 to 1 B stays at 1 C decreases from 1 to 0 D is undefined
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decreases from 1 to 0

The adjacent side shrinks as the angle opens.

Q81. The cosine row of the table is

Intermediate
A the same as the sine row B the sine row reversed C the sine row doubled D unrelated to the sine row
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the sine row reversed

That is why only one row needs memorising.

Q82. $\operatorname{cosec} 30°$ equals

Intermediate
A $\frac12$ B $\frac{2}{\sqrt3}$ C undefined D 2
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2

The reciprocal of $\sin 30° = \frac12$.

Q83. $\sec 60°$ equals

Intermediate
A $\frac12$ B $\frac{2}{\sqrt3}$ C 2 D $\sqrt3$
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2

The reciprocal of $\cos 60° = \frac12$.

Q84. $\cot 45°$ equals

Intermediate
A 0 B $\sqrt3$ C 1 D undefined
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1

The reciprocal of $\tan 45° = 1$.

Q85. $\operatorname{cosec} 0°$ is

Intermediate
A 0 B undefined C 1 D infinity
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undefined

It is $\frac{1}{\sin 0°} = \frac10$, which has no value.

Q86. $\operatorname{cosec} 30° + \sec 60°$ equals

Intermediate
A 2 B 4 C $\frac12$ D 3
Show answer & explanation

4

Both reciprocals are 2.

Q87. $\sin^2 60° + \cos^2 30°$ equals

Intermediate
A 1 B $\frac34$ C $\frac32$ D 0
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$\frac32$

Both squares are $\frac34$; note this is NOT the identity, which needs the same angle twice.

Q88. Assertion (A): $\tan 90°$ equals infinity. Reason (R): The adjacent side becomes zero as the angle approaches 90°.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

R is true, but division by zero has no value at all, so $\tan 90°$ is undefined rather than infinite — A is false.

Q89. Assertion (A): $\sin 60^\circ = \cos 60^\circ$. Reason (R): The cosine row of the table is the sine row read backwards.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

Those two are not equal; $\sin 60^\circ$ equals $\cos 30^\circ$. The assertion is false and the reason explains the pairing.

Q90. $\cot 30°$ equals

Intermediate
A $\sqrt3$ B $\frac{1}{\sqrt3}$ C 1 D $\frac12$
Show answer & explanation

$\sqrt3$

The reciprocal of $\tan 30° = \frac1{\sqrt3}$.

Q91. $\sec 45°$ equals

Intermediate
A $\frac1{\sqrt2}$ B $\sqrt2$ C 1 D 2
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$\sqrt2$

The reciprocal of $\cos 45° = \frac1{\sqrt2}$.

Q92. $\frac{\sin 60°}{\cos 60°}$ equals

Intermediate
A $\sqrt3$ B $\frac1{\sqrt3}$ C 1 D $\frac{\sqrt3}2$
Show answer & explanation

$\sqrt3$

$\frac{\sqrt3/2}{1/2} = \sqrt3$, which is $\tan 60°$.

Q93. As the acute angle increases, $\tan$

Intermediate
A decreases to zero B stays at 1 C is always undefined D increases without limit
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increases without limit

The opposite side grows while the adjacent side shrinks, so the ratio grows without bound.

Q94. Derive the values of $\sin 45°$, $\cos 45°$ and $\tan 45°$.

Intermediate
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Q95. Explain why $\tan 90°$ is undefined.

Intermediate
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Q96. Evaluate $\sin 30° \cos 60° + \cos 30° \sin 60°$.

Intermediate
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Q97. Evaluate $2 \tan^2 45° + \cos^2 30° - \sin^2 60°$.

Intermediate
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Q98. Evaluate $\cos 60° \cos 30° - \sin 60° \sin 30°$.

Intermediate
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Q99. Evaluate $\frac{\tan 30°}{\tan 60°}$.

Intermediate
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Q100. (a) Write out the complete table of $\sin$, $\cos$ and $\tan$ for 0°, 30°, 45°, 60° and 90°. (b) State two patterns in the table that reduce what must be memorised. (c) Use the table to evaluate $\sin^2 30° + \cos^2 30°$ and $\sin^2 45° + \cos^2 45°$, and comment.

Intermediate
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2 marks — (a) the complete table. 1½ marks — (b) both patterns. 1½ marks — (c) both evaluations and the observation.

Q101. Does $\sin^2\theta + \cos^2\theta = 1$ imply $\sin\theta + \cos\theta = 1$?

Intermediate
A yes B yes, for acute angles only C no — at 45° the sum is $\sqrt2$ D only at 0°
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no — at 45° the sum is $\sqrt2$

Square roots do not distribute across a sum.

Q102. $\frac{\sec\theta}{\operatorname{cosec}\theta}$ equals

Intermediate
A $\cot\theta$ B 1 C $\sec\theta$ D $\tan\theta$
Show answer & explanation

$\tan\theta$

$\frac1{\cos\theta} \times \sin\theta$.

Q103. $\sin^4\theta - \cos^4\theta$ simplifies to

Intermediate
A 1 B 0 C $\sin^2\theta - \cos^2\theta$ D $\sin^2\theta + \cos^2\theta$
Show answer & explanation

$\sin^2\theta - \cos^2\theta$

A difference of squares, whose first factor is 1.

Q104. $\frac{1}{\sec\theta - \tan\theta}$ simplifies to

Intermediate
A $\sec\theta + \tan\theta$ B $\sec\theta - \tan\theta$ C 1 D $\tan\theta$
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$\sec\theta + \tan\theta$

Multiply above and below by the conjugate; the denominator becomes 1.

Q105. The conjugate of $\operatorname{cosec}\theta + \cot\theta$ is

Intermediate
A $\operatorname{cosec}\theta + \cot\theta$ B $\cot\theta - \operatorname{cosec}\theta$ C $\operatorname{cosec}\theta - \cot\theta$ D $1 - \cot\theta$
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$\operatorname{cosec}\theta - \cot\theta$

The conjugate flips the sign between the two terms.

Q106. When multiplying by a conjugate you must multiply

Intermediate
A both the numerator and the denominator B the numerator only C the denominator only D neither
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both the numerator and the denominator

Otherwise the value of the expression changes.

Q107. $\cot\theta \times \sin\theta$ equals

Intermediate
A $\sin\theta$ B $\cos\theta$ C $\tan\theta$ D 1
Show answer & explanation

$\cos\theta$

$\frac{\cos\theta}{\sin\theta} \times \sin\theta$.

Q108. $\tan\theta + \cot\theta$ equals

Intermediate
A 1 B $\sec\theta \operatorname{cosec}\theta$ C $\sec\theta + \operatorname{cosec}\theta$ D 0
Show answer & explanation

$\sec\theta \operatorname{cosec}\theta$

Combining the fractions gives $\frac{\sin^2 + \cos^2}{\sin\cos} = \frac1{\sin\cos}$.

Q109. The most reliable move when a proof stalls is to

Intermediate
A square both sides B substitute 45° C give up and state the answer D convert everything to sin and cos
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convert everything to sin and cos

It puts every term into one common language.

Q110. Identity 2 is NOT valid at

Intermediate
A 0° B 45° C 90° D 30°
Show answer & explanation

90°

Both $\sec 90°$ and $\tan 90°$ are undefined.

Q111. Identity 3 is NOT valid at

Intermediate
A 90° B 0° C 60° D 45°
Show answer & explanation

0°

Both $\operatorname{cosec} 0°$ and $\cot 0°$ are undefined.

Q112. Assertion (A): All three identities are really the same fact. Reason (R): Identities 2 and 3 are obtained from identity 1 by division alone.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

No new information is added, so all three carry the content of Pythagoras.

Q113. Assertion (A): An identity can be solved to find $\theta$. Reason (R): An identity holds for every value of $\theta$.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true and is exactly why A is false — there is nothing to solve when every angle works.

Q114. $\cos\theta \times \operatorname{cosec}\theta$ equals

Intermediate
A $\tan\theta$ B 1 C $\cot\theta$ D $\sec\theta$
Show answer & explanation

$\cot\theta$

$\cos\theta \times \frac1{\sin\theta}$.

Q115. $\cos^2\theta \sec^2\theta + \sin^2\theta \operatorname{cosec}^2\theta$ equals

Intermediate
A 1 B 2 C 0 D $\tan^2\theta$
Show answer & explanation

2

Each product is 1, so the sum is 2.

Q116. $\tan\theta \cos\theta$ equals

Intermediate
A $\cos\theta$ B $\sin\theta$ C 1 D $\sec\theta$
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$\sin\theta$

$\frac{\sin\theta}{\cos\theta} \times \cos\theta$.

Q117. Assertion (A): Identity 1 can be rearranged into six different useful forms. Reason (R): Each identity can be written with either squared term isolated on one side.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Rearranging is what makes the identities usable inside longer expressions.

Q118. Prove that $\sin^2\theta + \cos^2\theta = 1$.

Intermediate
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Q119. Derive $\sec^2\theta - \tan^2\theta = 1$ from the first identity.

Intermediate
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Q120. Explain the difference between a trigonometric identity and a trigonometric equation.

Intermediate
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Q121. Prove that $(1 - \cos^2\theta)\operatorname{cosec}^2\theta = 1$.

Intermediate
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Q122. Simplify $\sin^4\theta - \cos^4\theta$.

Intermediate
Show answer & explanation

—

Topper level · 51 questions

Q123. If $\cot A = \frac{7}{8}$, then $\operatorname{cosec} A$ is

Advanced
A $\frac{\sqrt{113}}{8}$ B $\frac{8}{7}$ C $\frac{\sqrt{113}}{7}$ D $\frac{7}{\sqrt{113}}$
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$\frac{\sqrt{113}}{8}$

Adjacent 7 and opposite 8 give hypotenuse $\sqrt{49 + 64} = \sqrt{113}$, and cosec is hypotenuse over opposite.

Q124. Where $\sin A = \frac{8}{17}$, the tangent of A is

Advanced
A $\frac{15}{8}$ B $\frac{8}{15}$ C $\frac{8}{17}$ D $\frac{17}{15}$
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$\frac{8}{15}$

The adjacent side is $\sqrt{289 - 64} = 15$.

Q125. If $15\cot A = 8$, then $\sin A$ is

Advanced
A $\frac{8}{17}$ B $\frac{15}{8}$ C $\frac{15}{17}$ D $\frac{17}{15}$
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$\frac{15}{17}$

$\cot A = \frac{8}{15}$, so adjacent 8 and opposite 15 give hypotenuse 17.

Q126. The product $\sin A \times \operatorname{cosec} A$ equals

Advanced
A 1 B 0 C $\sin^2 A$ D $\tan A$
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1

A number times its reciprocal is 1.

Q127. The product $\tan A \times \cot A$ equals

Advanced
A 0 B 1 C $\tan^2 A$ D $\sin A$
Show answer & explanation

1

They are reciprocals of each other.

Q128. If $\sin A = \cos A$ for an acute angle A, the triangle is

Advanced
A equilateral B scalene C impossible D isosceles, with the two legs equal
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isosceles, with the two legs equal

Equal ratios over the same hypotenuse force the opposite and adjacent sides to be equal.

Q129. In a right triangle, $\frac{\sin A}{\cos A} \times \frac{\cos A}{\sin A}$ equals

Advanced
A 1 B $\tan^2 A$ C 0 D $\sin A \cos A$
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1

It is $\tan A \times \cot A$.

Q130. Which statement about an acute angle A is FALSE?

Advanced
A $\sin A \le 1$ B $\sec A \ge 1$ C $\operatorname{cosec} A$ can be less than 1 D $\tan A$ can exceed 1
Show answer & explanation

$\operatorname{cosec} A$ can be less than 1

Cosec is the reciprocal of something at most 1, so it is at least 1.

Q131. Assertion (A): Knowing one trigonometric ratio determines all the others. Reason (R): The given ratio fixes two sides of a right triangle, and Pythagoras then fixes the third.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Once all three sides are known in proportion, every ratio can be read off.

Q132. If the sides of a right triangle are doubled, the value of $\tan A$

Advanced
A doubles B halves C is unchanged D is squared
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is unchanged

Both sides in the ratio double, so the fraction is the same.

Q133. Explain the difference between $\sin^{-1} A$ and $\operatorname{cosec} A$.

Advanced
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Q134. If $15 \cot A = 8$, find $\sin A$ and $\sec A$.

Advanced
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Q135. (a) Explain why the trigonometric ratios of an angle do not depend on the size of the triangle. (b) Hence explain why, given $\tan A = \frac34$, you may take the opposite side as 3 and the adjacent as 4. (c) Using those sides, find $\sin A$, $\cos A$ and $\operatorname{cosec} A$.

Advanced
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2 marks — (a) the similarity argument. 1½ marks — (b) the 'any similar triangle' reasoning. 1½ marks — (c) all three values.

Q136. (a) State the limits on the values of $\sin A$, $\cos A$, $\sec A$ and $\operatorname{cosec} A$ for an acute angle, giving a reason for each. (b) Explain why $\tan A$ has no upper limit. (c) A student writes $\sin A = \frac{7}{5}$. Explain what is wrong and how it could be spotted immediately.

Advanced
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2 marks — (a) all four limits with reasons. 1½ marks — (b) the absence of the hypotenuse. 1½ marks — (c) the impossibility and how to notice it.

Q137. In a right triangle PQR, right-angled at Q, $PQ = 5$ cm and $PR = 13$ cm. (a) Find QR. (b) Find $\sin P$, $\cos P$ and $\tan P$. (c) Find $\sin R$ and $\cos R$, and comment on what you notice.

Advanced
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1 mark — (a) QR = 12. 2 marks — (b) all three ratios with the sides correctly identified. 2 marks — (c) both ratios and the observation about the swap.

Q138. $2 \tan^2 45° + \cos^2 30° - \sin^2 60°$ equals

Advanced
A 2 B 3 C $\frac12$ D 0
Show answer & explanation

2

$2(1) + \frac34 - \frac34 = 2$, since $\cos 30° = \sin 60°$.

Q139. $\sin(30° + 60°)$ equals

Advanced
A $\frac12 + \frac{\sqrt3}{2}$ B 1 C $\frac{\sqrt3}{2}$ D 0
Show answer & explanation

1

Evaluate the bracket first: $\sin 90° = 1$. It is NOT the sum of the two sines.

Q140. Which statement is true?

Advanced
A $\sin(A+B) = \sin A + \sin B$ always B $\sin(A+B) = \sin A \sin B$ C $\sin(A+B)$ is undefined D $\sin(A+B) \ne \sin A + \sin B$ in general
Show answer & explanation

$\sin(A+B) \ne \sin A + \sin B$ in general

$\sin 90° = 1$ while $\sin 30° + \sin 60° \approx 1.37$.

Q141. $\frac{\tan 30°}{\tan 60°}$ equals

Advanced
A 1 B $\frac13$ C 3 D $\frac{1}{\sqrt3}$
Show answer & explanation

$\frac13$

$\frac{1/\sqrt3}{\sqrt3} = \frac13$.

Q142. $\tan 60° - \cot 30°$ equals

Advanced
A 0 B $\sqrt3$ C $\frac{2}{\sqrt3}$ D 1
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0

Both terms are $\sqrt3$.

Q143. $2 \sin 30° \cos 30°$ equals

Advanced
A $\sin 60°$ B $\sin 30°$ C 1 D $\cos 60°$
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$\sin 60°$

$2 \cdot \frac12 \cdot \frac{\sqrt3}2 = \frac{\sqrt3}2 = \sin 60°$.

Q144. If $\sin \theta = \frac{\sqrt3}{2}$ for an acute angle, then $\theta$ is

Advanced
A 30° B 60° C 45° D 90°
Show answer & explanation

60°

Reading the sine row backwards identifies the angle.

Q145. If $\tan \theta = 1$ for an acute angle, then $\theta$ is

Advanced
A 30° B 45° C 60° D 0°
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45°

Only at 45° are the two legs equal.

Q146. Assertion (A): Every value in the standard table can be obtained from just two triangles. Reason (R): Half a square gives the $30^\circ$ and $60^\circ$ values, and half an equilateral triangle gives the $45^\circ$ values.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is true but R is false

The two triangles are the other way round. The assertion stands, the reason is false.

Q147. $\tan 45° \div \tan 60°$ equals

Advanced
A $\sqrt3$ B 1 C 3 D $\frac1{\sqrt3}$
Show answer & explanation

$\frac1{\sqrt3}$

$1 \div \sqrt3$.

Q148. Derive the values of $\sin 30°$ and $\cos 30°$.

Advanced
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—

Q149. Show that $\sin(30° + 60°) \ne \sin 30° + \sin 60°$.

Advanced
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—

Q150. List the standard values that are undefined, with a reason.

Advanced
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Q151. (a) Draw and describe the triangle used to obtain the 45° ratios, and state all three. (b) Do the same for the 30° and 60° ratios. (c) Explain why the 0° and 90° values cannot be obtained from a drawn triangle.

Advanced
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1½ marks — (a) the square construction and all three ratios. 2 marks — (b) the equilateral construction and both sets. 1½ marks — (c) the degeneracy argument and the limiting values.

Q152. (a) Evaluate $\frac{5 \cos^2 60° + 4 \sec^2 30° - \tan^2 45°}{\sin^2 30° + \cos^2 30°}$. (b) Explain why the denominator could have been written down without calculation. (c) State which standard values are undefined and why.

Advanced
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2½ marks — (a) each term evaluated and the fraction simplified. 1 mark — (b) the identity recognised. 1½ marks — (c) all the undefined values with the reason.

Q153. A surveyor uses standard angles for quick estimates. (a) A ramp makes 30° with the ground and is 20 m long. Find its vertical rise. (b) A roof makes 45° with the horizontal and rises 4 m. Find its horizontal run. (c) A wire makes 60° with the ground and is anchored 5 m from the base of a pole. Find the length of the wire.

Advanced
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1½ marks — (a) the correct ratio chosen and rise = 10 m. 1½ marks — (b) run = 4 m with the comment. 2 marks — (c) the correct ratio and wire = 10 m.

Q154. $\frac{1 - \cos\theta}{\sin\theta}$ equals

Advanced
A $\frac{\sin\theta}{1 + \cos\theta}$ B $\frac{\sin\theta}{1 - \cos\theta}$ C $\frac{1 + \cos\theta}{\sin\theta}$ D 1
Show answer & explanation

$\frac{\sin\theta}{1 + \cos\theta}$

Multiply top and bottom by $1 + \cos\theta$; the numerator becomes $\sin^2\theta$.

Q155. $\sec^2\theta + \operatorname{cosec}^2\theta$ equals

Advanced
A 1 B $\tan^2\theta + \cot^2\theta$ C $\sec^2\theta \operatorname{cosec}^2\theta$ D 2
Show answer & explanation

$\sec^2\theta \operatorname{cosec}^2\theta$

$\frac1{\cos^2} + \frac1{\sin^2} = \frac{\sin^2 + \cos^2}{\sin^2\cos^2} = \frac1{\sin^2\cos^2}$.

Q156. $1 - 2\sin^2\theta$ equals

Advanced
A $\cos^2\theta$ B $-\sin^2\theta$ C $\cos^2\theta - \sin^2\theta$ D 1
Show answer & explanation

$\cos^2\theta - \sin^2\theta$

Expanding gives $\sin^2 + 2\sin\cos + \cos^2$, and the outer terms make 1.

Q157. $\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta - 1$ equals

Advanced
A 1 B 2 C $-1$ D 0
Show answer & explanation

0

The first three terms make $(\sin^2\theta + \cos^2\theta)^2 = 1$.

Q158. If $\sin\theta = \frac35$, then $\cos^2\theta$ equals

Advanced
A $\frac9{25}$ B $\frac45$ C $\frac{25}{16}$ D $\frac{16}{25}$
Show answer & explanation

$\frac{16}{25}$

$\cos^2 = 1 - \frac9{25}$.

Q159. $\frac{\cos^2\theta}{1 - \sin\theta}$ equals

Advanced
A $1 - \sin\theta$ B $1 + \sin\theta$ C $\cos\theta$ D $\sin\theta$
Show answer & explanation

$1 + \sin\theta$

$\cos^2 = (1-\sin)(1+\sin)$, and the $(1-\sin)$ cancels.

Q160. $\sqrt{\sec^2\theta - 1}$ for an acute angle equals

Advanced
A $\sec\theta$ B $\tan\theta$ C $\cot\theta$ D 1
Show answer & explanation

$\tan\theta$

Identity 2 makes the inside $\tan^2\theta$, and tan is positive for an acute angle.

Q161. $\tan^2\theta - \sin^2\theta$ equals

Advanced
A 1 B 0 C $\sec^2\theta$ D $\tan^2\theta \sin^2\theta$
Show answer & explanation

$\tan^2\theta \sin^2\theta$

$\frac{\sin^2}{\cos^2} - \sin^2 = \sin^2\left(\frac{1 - \cos^2}{\cos^2}\right) = \sin^2 \tan^2$.

Q162. Assertion (A): Multiplying by a conjugate does not change the value of an expression. Reason (R): The conjugate is used as a fraction over itself, which equals 1.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Multiplying by 1 in disguise is precisely why the technique is legitimate.

Q163. Which technique suits $\sin^4\theta - \cos^4\theta$?

Advanced
A multiplying by a conjugate B substituting 45° C squaring both sides D factorising as a difference of squares
Show answer & explanation

factorising as a difference of squares

The factors are $\sin^2 + \cos^2$ and $\sin^2 - \cos^2$.

Q164. Identity 1 says the point $(\cos\theta, \sin\theta)$ always lies on

Advanced
A a straight line B a parabola C a circle of radius 1 about the origin D the x-axis
Show answer & explanation

a circle of radius 1 about the origin

$x^2 + y^2 = 1$ is exactly the unit circle.

Q165. Prove that $\tan\theta + \cot\theta = \sec\theta \operatorname{cosec}\theta$.

Advanced
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Q166. Show that $\frac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta$.

Advanced
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—

Q167. Prove that $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2$.

Advanced
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Q168. Simplify $\frac{\cos^2\theta}{1 - \sin\theta}$.

Advanced
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Q169. Explain why you must work on only one side when proving an identity.

Advanced
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Q170. (a) Prove that $\sin^2\theta + \cos^2\theta = 1$, stating clearly what you use. (b) Derive the other two identities from it. (c) State the range of angles for which each of the three is valid, with a reason.

Advanced
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2 marks — (a) Pythagoras, the division and the conclusion. 2 marks — (b) both derivations. 1 mark — (c) the three ranges with reasons.

Q171. (a) State the rule for proving a trigonometric identity, and why it matters. (b) Prove that $\frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}$. (c) Name the technique you used and explain why it works.

Advanced
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1 mark — (a) the rule and its reason. 2½ marks — (b) the conjugate step, the identity and the conclusion. 1½ marks — (c) naming the technique and justifying it.

Q172. (a) Prove that $\tan\theta + \cot\theta = \sec\theta \operatorname{cosec}\theta$. (b) Prove that $\sec^2\theta + \operatorname{cosec}^2\theta = \sec^2\theta \operatorname{cosec}^2\theta$. (c) State the single technique both proofs share.

Advanced
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2 marks — (a) the full proof. 2 marks — (b) the full proof. 1 mark — (c) naming the shared technique.

Q173. (a) Given that $\sin\theta = \frac{3}{5}$ for an acute angle, find $\cos\theta$ and $\tan\theta$ using an identity rather than a triangle. (b) Verify your answer satisfies $\sec^2\theta - \tan^2\theta = 1$. (c) Explain why using the identity gives the same answer as drawing a triangle.

Advanced
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2 marks — (a) both values with the positivity noted. 1½ marks — (b) the verification. 1½ marks — (c) the explanation linking identity and Pythagoras.

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