Q1 1 mark MCQ p. 88
Given, cot θ = 3, what is the value of cos θ?
Reveal official answer
Correct option: 3 — 3/√10
Question Bank · Mathematics
Introduction to Trigonometry
Source: CBSE · Competency Focused Practice Questions, Mathematics (Volume 3), Grade 10, pp. 88–102. Questions and answers are reproduced verbatim from the official CBSE document.
Try the question yourself before revealing the answer — that's how marks stick.
Q2 1 mark MCQ p. 88
Given below is ΔPQR, right-angled at Q. (In the figure, θ is marked at vertex P.)(Note: The figure is not to scale.)What is the value of tan θ?
Reveal official answer
Correct option: 2 — QR/PQ
Q3 1 mark MCQ p. 88
Given that cos²θ − sin²θ = ¾, what is the value of cos θ?
Reveal official answer
Correct option: 2 — √7/2√2
Q4 1 mark MCQ p. 88
If cot 81° = tan θ, what is the value of sec 5θ?(Note: 0° ≤ 5θ ≤ 90°)
Reveal official answer
Correct option: 3 — √2
Q5 1 mark MCQ p. 89
Any relation which is ALWAYS true is an identity. Which of the following is a trigonometric identity?i) cot θ = cos θ/sin θii) sec²θ + cosec²θ = 1iii) (1 − cos²θ)/cos²θ = tan²θiv) sin²θ/(1 − sin²θ) = cot²θ
Reveal official answer
Correct option: 2 — only i) and iii)
Q7 1 mark MCQ p. 89
Which of these is equal to √[(1 + sin θ)/(1 − sin θ)]?
Reveal official answer
Correct option: 1 — sec θ + tan θ
Q8 1 mark MCQ p. 89
If cos θ = 12/13, what is the value of 5cosec θ − 4tan θ?
Reveal official answer
Correct option: 4 — 34/3
Q9 1 mark MCQ p. 90
Answer the questions based on the given information.A TRIG-QUIZ is organized in a school, which consists of 3 rounds in order to get a winner. 4 Teams participated in the quiz - Team 1, Team 2, Team 3 and Team 4.In each round, the same question was asked to all the teams and one team was eliminated after every round.Following three questions were asked in 3 rounds to the teams.Round 1: Give a correct statement related to trigonometric ratio of an angle Θ.Round 2: If cos²Θ − sin²Θ = ¾, 0° < Θ < 90°, use appropriate identities to find the values of cos Θ and tan Θ.Round 3: In a right angled triangle ABC, B is at right angle and sin A = ¼.Find the value of: cos²A + 2sin²A + 2sin²C + 2cos²CWhich team gets eliminated in Round 1 if the following are the statements made by the teams:Team 1: cos θ = −1, for some angle θ.Team 2: tan θ = 2, for some angle θ.Team 3: sin θ = 2, for some angle θ.Team 4: tan θ = 10, for some angle θ.
Reveal official answer
Correct option: 3 — Team 3
Q10 2 marks Written p. 90
Answer the questions based on the given information.A TRIG-QUIZ is organized in a school, which consists of 3 rounds in order to get a winner. 4 Teams participated in the quiz - Team 1, Team 2, Team 3 and Team 4.In each round, the same question was asked to all the teams and one team was eliminated after every round.Following three questions were asked in 3 rounds to the teams.Round 1: Give a correct statement related to trigonometric ratio of an angle Θ.Round 2: If cos²Θ − sin²Θ = ¾, 0° < Θ < 90°, use appropriate identities to find the values of cos Θ and tan Θ.Round 3: In a right angled triangle ABC, B is at right angle and sin A = ¼.Find the value of: cos²A + 2sin²A + 2sin²C + 2cos²CWhat answers should the teams give to enter Round 3. Show your work.
Reveal official answer
• Uses identity cos²θ + sin²θ = 1 and writes: 2cos²θ − 1 = ¾ [1]
• Simplifies the above and finds the values as: cos θ = √7/√8, tan θ = 1/√7 [1]
Q11 2 marks Written p. 90
Answer the questions based on the given information.A TRIG-QUIZ is organized in a school, which consists of 3 rounds in order to get a winner. 4 Teams participated in the quiz - Team 1, Team 2, Team 3 and Team 4.In each round, the same question was asked to all the teams and one team was eliminated after every round.Following three questions were asked in 3 rounds to the teams.Round 1: Give a correct statement related to trigonometric ratio of an angle Θ.Round 2: If cos²Θ − sin²Θ = ¾, 0° < Θ < 90°, use appropriate identities to find the values of cos Θ and tan Θ.Round 3: In a right angled triangle ABC, B is at right angle and sin A = ¼.Find the value of: cos²A + 2sin²A + 2sin²C + 2cos²CThe remaining teams were asked the Round 3 question.What answer should a team give to win the Quiz? Show your work.
Reveal official answer
• Uses the given equation sin A = ¼ to find other ratios as: cos A = √15/4, sin C = √15/4, cos C = ¼ [1]
• Substitutes the above values in the given expression, cos²A + 2sin²A + 2sin²C + 2cos²C and simplifies it to get 49/16. [1]
Q12 1 mark Written p. 90
If tan x − cot y = 0, find the value of x + y.Show your steps.(Note: 0° ≤ x, y ≤ 90°.)
Reveal official answer
• Rewrites that the given equation as tan x = cot y. [0.5]
• Concludes that this is only possible for a pair of complementary angles. Hence, x + y = 90°. [0.5]
Q13 1 mark Written p. 90
Prove that:sin²70° − sin²10° = cos²10° − cos²70°
Reveal official answer
• Proceeds with LHS as, = sin²70° − sin²10°. Uses identity sin²θ = 1 − cos²θ to get = (1 − cos²70°) − (1 − cos²10°) = 1 − cos²70° − 1 + cos²10° = cos²10° − cos²70° = RHS [1]
Q14 1 mark Written p. 90
Write true or false for the given statement and give a valid reason.In ΔABC, right-angled at B, cosec A can be less than 1.
Reveal official answer
• Writes False. [0.5]
• Gives a reason. For example, writes that as, in a right-angled triangle, cosec A is always greater than or equal to 1, as it represents the reciprocal of the sine function, and the sine of an acute angle is always less than or equal to 1. [0.5]
Q15 1 mark Written p. 91
What is the value of sin²θ + sec²θ + cos²θ − tan²θ where θ is an acute angle?Show your work.
Reveal official answer
• Rearranges the given expression and uses identities to evaluate as: (sin²θ + cos²θ) + (sec²θ − tan²θ) = 1 + 1 = 2 [1]
Q16 2 marks Written p. 91
Find the value of θ for which the below statement is true. θ is acute angle.√3tan θ − cot 45° = 0Show your work.
Reveal official answer
• Simplifies the given equation as tan θ = 1/√3. [1]
• Finds the value of θ for which tan θ is 1/√3 as 30°. [1]
Q17 2 marks Written p. 91
In ΔABC, AC = 25 cm and sin C = ⅘.Find the length of BC. Show your work.
Reveal official answer
• Writes: sin C = AB/AC = ⅘. [0.5]
• Substitutes the value of AC as 25 in the above equation and simplifies it to find the value of AB as 20 cm. [0.5]
• Uses Pythagoras theorem to find the length of BC as: √(25² − 20²) = 15 cm. [1]
Q18 2 marks Written p. 91
sin (A + B) = √3/2 and sin (A − B) = ½ where A and B are acute angles.Find the values of A and B. Show your steps.
Reveal official answer
• Writes that, since sin (A + B) = √3/2, A + B = 60°. [0.5]
• Writes that, since sin (A − B) = ½, A − B = 30°. [0.5]
• Solves the equations in steps 1 and 2 to find A as 45° and B as 15°. [1]
Q19 2 marks Written p. 91
2sin 3A = √3 where 3A is an acute angle.Find the value of A. Show your steps.
Reveal official answer
• Rewrites the above equation as: sin 3A = √3/2 [0.5]
• From the above step finds 3A as: sin 3A = √3/2 => sin 3A = sin 60° => 3A = 60° [1]
• Thus finds the value of A as 60/3 = 20°. [0.5]
Q20 2 marks Written p. 92
Shown below is a glass prism. When a ray of light enters the prism, it refracts inside the prism as shown.If the refractive index (RI) of the above prism is sec 45° and the angle of refraction (R) is 30°, find the angle of incidence (I). Show your work.(Note: Refractive Index = sin I/sin R.)
Reveal official answer
• Writes the equation for the refractive index as: Refractive Index = sin I/sin R => sec 45° = sin I/sin 30° => √2 = sin I/0.5 => sin I = 1/√2 [1]
• Finds the value of I for which sin I is 1/√2, that is 45°. [1]
Q21 2 marks Written p. 92
In the figure below, ABCD is a rectangle. (In the figure, the diagonal AC is drawn; ∠DAC = 60° and DC = 12 cm.)(Note: The figure is not to scale.)Find the length of BC and AC. Show your work.
Reveal official answer
• Writes that in ΔADC, tan 60° = 12/AD => AD = 4√3 cm. Finds BC = AD = 4√3 cm. [1]
• Uses Pythagoras theorem in ΔADC to find AC as √[12² + (4√3)²] = 8√3 cm. (Award full marks if the AC is found using other trigonometric ratios.) [1]
Q22 3 marks Written p. 93
In a ΔABC, ∠C is 37° and CB = 20 cm as shown below. (In the figure, the right angle is at B.)(Note: The figure is not to scale.)Findi) Length of AB.ii) sin 37°Show your work.(Note: Take cos 37° as 0.8.)
Reveal official answer
• i) Writes, in the ΔABC: cos C = CB/CA => cos 37° = 20/CA => CA = 20/0.8 = 25 cm. [1]
• Uses Pythagoras theorem to find AB as: AB = √(25² − 20²) = 15 cm. [1]
• ii) Finds sin 37° as: sin C = AB/AC => sin 37° = 15/25 = 0.6. (Award equal marks if a student uses an alternative method.) [1]
Q23 3 marks Written p. 93
Prove that:√[(sec θ − 1)/(sec θ + 1)] = cosec θ − cot θ
Reveal official answer
• Multiplies both numerator and denominator with (sec θ − 1) as: √{[(sec θ − 1)(sec θ − 1)] / [(sec θ + 1)(sec θ − 1)]} [1]
• Uses the identity sec²θ − 1 = tan²θ in the above expression as: = √[(sec θ − 1)²/(sec²θ − 1)] = √[(sec θ − 1)²/tan²θ] = (sec θ − 1)/tan θ [1.5]
• Further simplifies the above expression as: = sec θ/tan θ − 1/tan θ = cosec θ − cot θ. Hence, proves the given statement. [0.5]
Q24 3 marks Written p. 93
A ΔPQR is right angled at Q. If tan P = √5 − 2, show that sin P × cos P = 1/2√5. Show your work.
Reveal official answer
• Finds the hypotenuse as: √[(√5 − 2)² + 1²] = √(10 − 4√5) units [1]
• Finds the value of sin P and cos P as (√5 − 2)/√(10 − 4√5) and 1/√(10 − 4√5) respectively. [1]
• Calculates the value of sin P × cos P as (√5 − 2)/(10 − 4√5) and simplifies it further as 1/2√5. [1]
Q25 3 marks Written p. 93
Prove:tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θShow your work.
Reveal official answer
• Rewrites the LHS of the above equation as: (sin θ/cos θ)/(1 − cos θ/sin θ) + (cos θ/sin θ)/(1 − sin θ/cos θ). Simplifies the above expression as: [1/(sin θ − cos θ)] × [(sin³θ − cos³θ)/(sin θ cos θ)]. [1]
• Uses the formula (a³ − b³) = (a − b)(a² + ab + b²) in the above expression and simplifies it as: (sin²θ + cos²θ + sin θ cos θ)/(sin θ cos θ). [1]
• Further simplifies the above expression as: (1 + sin θ cos θ)/(sin θ cos θ). Replaces 1/sin θ with cosec θ and 1/cos θ with sec θ in the above expression and simplifies it as: 1 + sec θ cosec θ. Hence, proves the given statement. [1]
Q26 5 marks Written p. 93
Prove that:(tan θ + sec θ − 1)/(tan θ − sec θ + 1) = (1 + sin θ)/cos θ
Reveal official answer
• Uses identity sec²θ − tan²θ = 1 in LHS of the above equation and rewrites it as: [(tan θ + sec θ) − (sec²θ − tan²θ)] / (tan θ − sec θ + 1) [1]
• Rewrites the above equation as: [(tan θ + sec θ) − (sec θ + tan θ)(sec θ − tan θ)] / (tan θ − sec θ + 1) [1]
• Takes (tan θ + sec θ) common in the numerator as: [(tan θ + sec θ)(1 − sec θ + tan θ)] / (tan θ − sec θ + 1) [1]
• Rearrange the numerator and simplifies the above expression as: [(tan θ + sec θ)(tan θ − sec θ + 1)] / (tan θ − sec θ + 1) = tan θ + sec θ = sin θ/cos θ + 1/cos θ [1]
• Further simplifies the above expression and proves: (1 + sin θ)/cos θ = RHS [1]
Q27 5 marks Written p. 94
Solve the following:i) Given tan A = 5/12, find sin A, cos A, cot A, sec A, cosec A.ii) Given 4cos²A + 8sin²A = 5, show that cot A = √3.Show your work.
Reveal official answer
• i) Uses the pythagoras theorem and finds the hypotenuse as √(144 + 25) = 13 units. [1]
• Determines the other ratios as: sin A = 5/13, cos A = 12/13, cot A = 12/5, sec A = 13/12, cosec A = 13/5 [2]
• ii) Divides the given equation with sin²A to get: 4cot²A + 8 = 5cosec²A. [1]
• Uses identity, cosec²θ = 1 + cot²θ in RHS to get: 4cot²A + 8 = 5(1 + cot²A). Simplifies the above to get cot A = √3. Thus shows that, cot A = √3. [1]