CBSE Class 10 Mathematics Trigonometry Assignment Set 02

Multiple Choice Questions

Question. If 2 sin θ = √3 , then θ =
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Answer : B

Question. The value of (1 + tan² θ)(1 – sin θ)(1 + sin θ) is
(a) 0
(b) 1
(c) 8
(d) 17
Answer : B

Question. If 4 tanθ = 3, then [4sinθ - cosθ / 4sinθ + cosθ] is equal to
(a) 2/3
(b) 1/3
(c) 1/2
(d) 3/4
Answer : C

Question. If cosec θ + cot θ = x, the value of cosec θ – cot θ is
(a) x
(b) 2x
(c) x/2
(d) 1/x
Answer : D

Question. If sin θ – cos θ = 0, then the value of (sin4 θ + cos4 θ) is
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4
Answer : C

Question. The value of cot² θ – 1/2 sin²θ is
(a) 0
(b) –1
(c) 2
(d) –8
Answer : B

Question. The magnitude of θ in the equation cos²θ /cot²θ  - cos²θ = 3 is
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer : C

Question. The value of (1 + cotθ – cosecθ) (1 + tanθ + secθ) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Answer : B

Question. If 7 sin²A + 3 cos²A = 4, then tan A =
(a) 1/2
(b) 1/3
(c) 1/√2
(d) 1/√3
Answer : D

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): sin² 67° + cos² 67° = 1.
Reason (R): For any value of θ, sin² θ + cos²θ = 1.
Answer : A

Question. Assertion (A): The value of sec² 10° – cot² 80° is 1.
Reason (R): The value of sin 30° = 1/2
Answer : B

Case Study Based Questions

Two aeroplanes leave an airport, one after the other. After moving on runway, one flies due North and other flies due South. The speed of two aeroplanes are 400 km/hr and 500 km/hr respectively. Considering PQ as runway and A and B are any points in the path followed by two planes

On the basis of above answer the following questions

Question. Find tan𝜃, if ∠𝐴𝑃𝑄 = 𝜃
(a) 3/4
(b) 1/2
(c) 1/√2
(d) √3/2
Answer : A

Question. Find the value of cot𝐵
(a) 3/4
(b) 15/4
(c) 3/8
(b) 15/8
Answer : D

Question. Find the value of tan𝐴
(a) 3/4
(b) 4/3
(c) 1/√2
(d) √3/2
Answer : B

Question. Find the value of sec A
(a) 0
(b) 5/3
(c) 1/√2
(d) √3/2
Answer : B

Question. Find cosec B
(a) 17/8
(b) 8/17
(c) 12/5
(d) 5/12
Answer : A

Level 1 (1 Mark)

Question. What is the Value of \( \sin^2 A + \cos^2 A \)?
Answer: \( \sin^2 A + \cos^2 A = 1 \)

Question. What is the Value of \( \tan(90^\circ - A) \)?
Answer: \( \cot A \)

Question. If \( \tan A = \cot B \), What is the Value of \( A + B \)?
Answer:
\( \tan A = \cot A \)
\( \tan A = \tan(90^\circ - B) \)
\( A = 90^\circ - B \)
\( A + B = 90^\circ \)

Question. What is the Value of \( \sec 30^\circ \)?
Answer: \( \frac{2}{\sqrt{3}} \)

Level 2 (2 Marks)

Question. Evaluate \( \cos 60^\circ \sin 30^\circ + \sin 60^\circ + \cos 30^\circ \)
Answer:
\( \cos 60^\circ \sin 30^\circ + \sin 60^\circ + \cos 30^\circ \)
\( = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \)
\( = \frac{1}{4} + \frac{3}{4} = \frac{1 + 3}{4} = \frac{4}{4} = 1 \)

Question. If \( \sec^2 A (1 + \sin A)(1 - \sin A) = k \), find the value of \( k \).
Answer:
\( \sec^2 A (1 - \sin^2 A) = k \)
\( \sec^2 A (\cos^2 A) = k \)
\( \left(\frac{1}{\cos^2 A}\right) \cos^2 A = k \)
\( k = 1 \)

Question. If \( \sin A = \frac{1}{3} \), then find the value of \( (2 \cot^2 A + 2) \).
Answer:
\( 2 \cot^2 A + 2 = 2 (\dots^2 A + 1) \)
\( = 2 (\csc^2 A) \)
\( = 2 \left(\frac{1}{\sin^2 A}\right) \)
\( = \frac{2}{\left(\frac{1}{3}\right)^2} = \frac{2}{1/9} = \frac{2 \times 9}{1} = 18 \)

Question. If \( \tan A = \cot B \), prove that \( A + B = 90^\circ \)
Answer:
\( \tan A = \cot A \)
\( \tan A = \tan(90^\circ - B) \)
\( A = 90^\circ - B \)
\( A + B = 90^\circ \)

Level 3 (3 Marks)

Question. If \( \cot A = \frac{7}{8} \) then what is the value of \( \frac{(1 + \cos A)(1 - \cos A)}{(1 - \sin A)(1 + \sin A)} \)?
Answer:
\( \frac{(1 + \cos A)(1 - \cos A)}{(1 - \sin A)(1 + \sin A)} \)
\( \frac{\sin^2 A}{\cos^2 A} = \tan^2 A = \frac{1}{\cot^2 A} \)
\( \tan^2 A = \frac{1}{\cot^2 A} = \frac{1}{(7/8)^2} \)
\( = \left(\frac{8}{7}\right)^2 \)
\( = \frac{64}{49} \)

Question. Write the value of \( 2 \cos^2 A + \frac{2}{1 + \cot^2 A} \)
Answer:
\( 2 \cos^2 A + \frac{2}{1 + \cot^2 A} \)
\( = 2 \cos^2 A + \frac{2}{\csc^2 A} \)
\( = 2 \cos^2 A + 2 \sin^2 A \)
\( = 2 (\cos^2 A + \sin^2 A) \)
\( = 2(1) = 2 \)

Question. Given that \( \tan A = \frac{12}{5} \), calculate \( \sin A \), \( \cos A \) and \( \sec A \).
Answer:
Let \( ABC \) be a triangle right angled at \( B \).
As \( \tan A = \frac{12}{5} \),
Let \( BC = 12k \), \( AB = 5k \).
Using Pythagoras Theorem, \( AC^2 = CB^2 + BA^2 \)
\( = (12k)^2 + (5k)^2 = 169k^2 \)
so, \( AC = 13k \)
\( \sin A = \frac{BC}{AC} = \frac{12k}{13k} = \frac{12}{13} \)
\( \cos A = \frac{AB}{AC} = \frac{5k}{13k} = \frac{5}{13} \)
and \( \sec A = \frac{1}{\cos A} = \frac{13k}{5} \)

Question. Prove that \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \)
Answer:
L.H.S = \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} \times \sqrt{\frac{1 + \sin A}{1 + \sin A}} \)
\( = \frac{\sqrt{(1 + \sin A)^2}}{\sqrt{1 - \sin^2 A}} \)
\( = \frac{1 + \sin A}{\cos A} \)
\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \)
\( = \sec A + \tan A = \) R.H.S

Level 4 (4 Marks)

Question. Prove that \( \frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} = 1 + \sin \theta \cos \theta \)
Answer:
L.H.S = \( \frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} \)
\( = \frac{\cos^2 \theta}{1 - \frac{\sin \theta}{\cos \theta}} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} \)
\( = \frac{\cos^3 \theta}{\cos \theta - \sin \theta} - \frac{\sin^3 \theta}{\cos \theta - \sin \theta} \)
\( = \frac{\cos^3 \theta - \sin^3 \theta}{\cos \theta - \sin \theta} \)
\( = \frac{(\cos \theta - \sin \theta)(\cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta)}{\cos \theta - \sin \theta} \)
\( = 1 + \sin \theta \cos \theta \)

Question. Prove that \( \frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta} \)
Answer:
L.H.S = \( \frac{\tan \theta - \sec \theta - 1}{\tan \theta - \sec \theta + 1} \)
\( = \frac{(\tan \theta - \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1} \)
\( = \frac{1 + \sec \theta}{\cos \theta} = \) R.H.S

Question. Prove that \( \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} + \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = 2 \csc \theta \)
Answer:
L.H.S = \( \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} \times \sqrt{\frac{1 + \cos \theta}{1 + \cos \theta}} + \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \times \sqrt{\frac{1 - \cos \theta}{1 - \cos \theta}} \)
\( = \sqrt{\frac{(1 + \cos \theta)^2}{1 - \cos^2 \theta}} + \sqrt{\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}} \)
\( = \frac{1 + \cos \theta}{\sin \theta} + \frac{1 - \cos \theta}{\sin \theta} \) (using \( 1 - \cos^2 \theta = \sin^2 \theta \))
\( = \frac{1 + \cos \theta + 1 - \cos \theta}{\sin \theta} \)
\( = \frac{2}{\sin \theta} = 2 \csc \theta \)

Question. If \( \sec \theta + \tan \theta = p \), prove that \( \sin \theta = \frac{p^2 - 1}{p^2 + 1} \)
Answer:
\( \sec \theta + \tan \theta = p \) ————— (i)
\( \sec^2 \theta - \tan^2 \theta = 1 \)
\( \Rightarrow (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \) ————— (ii)
Dividing (ii) by (i) we get:
\( \sec \theta - \tan \theta = \frac{1}{p} \) ————— (iii)
Adding (i) and (iii) we get:
\( \sec \theta - \tan \theta + \sec \theta + \tan \theta = p + \frac{1}{p} \)
\( \Rightarrow 2\sec \theta = \frac{1 + p^2}{p} \) ————— (iv)
Similarly, \( 2\tan \theta = \frac{p^2 - 1}{p} \) ————— (v)
Dividing (v) by (iv) we get:
\( \sin \theta = \frac{p^2 - 1}{p^2 + 1} \)