CBSE Class 10 Mathematics Trigonometry Assignment Set 31

Exercise-1 Objective: Trigonometric Ratios

A-1. If \( 5\tan\theta = 4 \), then the value of \( \dfrac{5\sin\theta - 3\cos\theta}{5\sin\theta + 2\cos\theta} \) is
(A) \( \dfrac{1}{3} \)   (B) \( \dfrac{1}{6} \)   (C) \( \dfrac{4}{5} \)   (D) \( \dfrac{2}{3} \)
Answer:
Ans. (B) \( \dfrac{1}{6} \)
Divide the numerator and the denominator by \( \cos\theta \): \( \dfrac{5\tan\theta - 3}{5\tan\theta + 2} = \dfrac{4 - 3}{4 + 2} = \dfrac{1}{6} \).
In simple words: 5 tan θ = 4 is used as a whole.

Teacher's Note:
1. Divide by cos θ to get tan θ.

A-2. If \( 7\sin\alpha = 24\cos\alpha \), where \( 0^\circ < \alpha < 90^\circ \), then the value of \( 14\tan\alpha - 75\cos\alpha - 7\sec\alpha \) is equal to
(A) 1   (B) 2   (C) 3   (D) 4
Answer:
Ans. (B) 2
\( \tan\alpha = \dfrac{24}{7} \), so the triangle is 7, 24, 25: \( \cos\alpha = \dfrac{7}{25} \), \( \sec\alpha = \dfrac{25}{7} \).
\( 14 \times \dfrac{24}{7} - 75 \times \dfrac{7}{25} - 7 \times \dfrac{25}{7} = 48 - 21 - 25 = 2 \).
In simple words: Read all ratios from one triangle.

Teacher's Note:
1. Build the 7-24-25 triangle.

A-3. If \( \tan\theta = 4 \), then \( \dfrac{\tan\theta}{\dfrac{\sin^3\theta}{\cos\theta} + \sin\theta\cos\theta} \) is equal to
(A) 0   (B) \( 2\sqrt{2} \)   (C) \( \sqrt{2} \)   (D) 1
Answer:
Ans. (D) 1
The denominator is \( \sin\theta\left(\dfrac{\sin^2\theta}{\cos\theta} + \cos\theta\right) = \sin\theta \cdot \dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta} = \tan\theta \).
So the expression is \( \dfrac{\tan\theta}{\tan\theta} = 1 \), whatever the value of tan θ.
In simple words: Use sin²θ + cos²θ = 1.

Teacher's Note:
1. The denominator simplifies to tan θ.

A-4. If A is an acute angle in a right \( \triangle ABC \), right angled at B, then the value of \( \sin A + \cos A \) is
(A) equal to one   (B) greater than one   (C) less than one   (D) equal to two
Answer:
Ans. (B) greater than one
\( \sin A + \cos A = \dfrac{BC + AB}{AC} \). The sum of two sides of a triangle is greater than the third, so the value is greater than 1.
In simple words: The triangle inequality settles it.

Teacher's Note:
1. Two sides of a triangle add up to more than the third.

Exercise-1 Objective: Trigonometric Angles

B-1. As x increases from 0° to 90°, the value of cos x
(A) increases   (B) decreases   (C) remains constant   (D) increases, then decreases
Answer:
Ans. (B) decreases
cos 0° = 1 and cos 90° = 0, and cosine keeps falling in between: it decreases.
In simple words: Check the end values 1 and 0.

Teacher's Note:
1. The cosine of an angle falls as the angle grows.

B-2. The value of x from the equation \( x\sin\dfrac{\pi}{6}\cos^2\dfrac{\pi}{4} = \dfrac{\cot^2\frac{\pi}{6}\sec\frac{\pi}{3}\tan\frac{\pi}{4}}{\operatorname{cosec}^2\frac{\pi}{4}\operatorname{cosec}\frac{\pi}{6}} \) is
(A) 4   (B) 6   (C) \( -2 \)   (D) 0
Answer:
Ans. (B) 6
Left side: \( x \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{x}{4} \). Right side: \( \dfrac{3 \times 2 \times 1}{2 \times 2} = \dfrac{3}{2} \).
\( \dfrac{x}{4} = \dfrac{3}{2} \Rightarrow x = 6 \).
In simple words: π/6, π/4, π/3 are 30°, 45°, 60°.

Teacher's Note:
1. Replace π/6 by 30°, π/4 by 45°, π/3 by 60°.

B-3. The area of a triangle is 12 sq. cm. Two sides are 6 cm and 12 cm. The included angle is \( \theta \), then \( \sin\theta = \)
(A) \( \dfrac{1}{2} \)   (B) \( \dfrac{1}{4} \)   (C) \( \dfrac{1}{6} \)   (D) \( \dfrac{1}{3} \)
Answer:
Ans. (D) \( \dfrac{1}{3} \)
Area \( = \dfrac{1}{2} \times 6 \times 12 \times \sin\theta = 36\sin\theta = 12 \Rightarrow \sin\theta = \dfrac{1}{3} \).
In simple words: The formula uses the angle between the two sides.

Teacher's Note:
1. Area = ½ ab sin θ.

B-4. If \( \alpha + \beta = 90^\circ \) and \( \alpha = 2\beta \), then \( \cos^2\alpha + \sin^2\beta \) equals
(A) \( \dfrac{1}{2} \)   (B) 0   (C) 1   (D) 2
Answer:
Ans. (A) \( \dfrac{1}{2} \)
\( 3\beta = 90^\circ \), so \( \beta = 30^\circ \) and \( \alpha = 60^\circ \). \( \cos^2 60^\circ + \sin^2 30^\circ = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \).
In simple words: α = 2β and α + β = 90°.

Teacher's Note:
1. Solve for the two angles first.

B-5. Given that \( \sin A = \dfrac{1}{2} \) and \( \cos B = \dfrac{1}{\sqrt{2}} \), where A and B are acute angles, then the value of A + B is
(A) 30°   (B) 45°   (C) 75°   (D) 15°
Answer:
Ans. (C) 75°
\( A = 30^\circ \) and \( B = 45^\circ \), so \( A + B = 75^\circ \).
In simple words: sin 30° = 1/2 and cos 45° = 1/√2.

Teacher's Note:
1. Read the angles from the standard table.

Exercise-1 Objective: Complementary Angles

C-1. If \( \alpha + \beta = 90^\circ \) and \( \sin\alpha = \dfrac{1}{3} \), then \( \sin\beta \) is
(A) \( \dfrac{\sqrt{2}}{3} \)   (B) \( \dfrac{2\sqrt{2}}{3} \)   (C) \( \dfrac{2}{3} \)   (D) \( \dfrac{3}{4} \)
Answer:
Ans. (B) \( \dfrac{2\sqrt{2}}{3} \)
\( \sin\beta = \cos\alpha = \sqrt{1 - \dfrac{1}{9}} = \dfrac{2\sqrt{2}}{3} \).
In simple words: Use sin²α + cos²α = 1.

Teacher's Note:
1. sin β = cos α for complementary angles.

C-2. The value of \( \tan 5^\circ \cdot \tan 10^\circ \cdot \tan 15^\circ \cdot \tan 20^\circ \cdots \tan 85^\circ \) is
(A) 1   (B) 2   (C) 3   (D) None of these
Answer:
Ans. (A) 1
Pair \( \tan\theta \) with \( \tan(90^\circ - \theta) = \cot\theta \): each pair gives 1, and \( \tan 45^\circ = 1 \). The product is 1.
In simple words: tan 45° = 1 sits in the middle.

Teacher's Note:
1. Each complementary pair multiplies to 1.

C-3. \( \sin(60^\circ + \theta) - \cos(30^\circ - \theta) \) is equal to
(A) \( 2\cos\theta \)   (B) \( 2\sin\theta \)   (C) 0   (D) 1
Answer:
Ans. (C) 0
\( \cos(30^\circ - \theta) = \sin(90^\circ - 30^\circ + \theta) = \sin(60^\circ + \theta) \), so the difference is 0.
In simple words: Convert the cosine to a sine.

Teacher's Note:
1. cos x = sin(90° − x).

C-4. If \( \cos(\alpha + \beta) = 0 \), then \( \sin(\alpha - \beta) \) can be reduced to
(A) \( \cos\beta \)   (B) \( \cos 2\beta \)   (C) \( \sin\alpha \)   (D) \( \sin 2\alpha \)
Answer:
Ans. (B) \( \cos 2\beta \)
\( \alpha + \beta = 90^\circ \Rightarrow \alpha = 90^\circ - \beta \). \( \sin(\alpha - \beta) = \sin(90^\circ - 2\beta) = \cos 2\beta \).
In simple words: Replace α by 90° − β.

Teacher's Note:
1. cos 90° = 0 gives α + β = 90°.

C-5. If \( \sec A = \operatorname{cosec} B = \dfrac{12}{7} \), then A + B is
(A) zero   (B) 90°   (C) < 90°   (D) > 90°
Answer:
Ans. (B) 90°
\( \sec A = \operatorname{cosec} B = \sec(90^\circ - B) \), so \( A = 90^\circ - B \) (both acute), that is \( A + B = 90^\circ \).
In simple words: Both angles are acute, so they are complementary.

Teacher's Note:
1. cosec B = sec(90° − B).

Exercise-1 Objective: Trigonometric Identities

D-1. Given \( 3\sin\beta + 5\cos\beta = 5 \), then the value of \( (3\cos\beta - 5\sin\beta)^2 \) is equal to
(A) 9   (B) \( \dfrac{9}{5} \)   (C) \( \dfrac{1}{3} \)   (D) \( \dfrac{1}{9} \)
Answer:
Ans. (A) 9
\( (3\sin\beta + 5\cos\beta)^2 + (3\cos\beta - 5\sin\beta)^2 = 9 + 25 = 34 \).
So \( (3\cos\beta - 5\sin\beta)^2 = 34 - 25 = 9 \).
In simple words: The cross terms cancel.

Teacher's Note:
1. Square and add the two expressions.

D-2. The value of \( [(\sec A + \tan A)(1 - \sin A)] \) is equal to
(A) \( \tan^2 A \)   (B) \( \sin^2 A \)   (C) \( \cos A \)   (D) \( \sin A \)
Answer:
Ans. (C) \( \cos A \)
\( \left(\dfrac{1 + \sin A}{\cos A}\right)(1 - \sin A) = \dfrac{1 - \sin^2 A}{\cos A} = \dfrac{\cos^2 A}{\cos A} = \cos A \).
In simple words: (1 + sin A)(1 − sin A) = cos²A.

Teacher's Note:
1. Write sec and tan with sin and cos.

D-3. If \( x = 2\sin^2\theta \) and \( y = 2\cos^2\theta + 1 \), then the value of x + y is
(A) 2   (B) 3   (C) \( \dfrac{1}{2} \)   (D) 1
Answer:
Ans. (B) 3
\( x + y = 2(\sin^2\theta + \cos^2\theta) + 1 = 3 \).
In simple words: sin²θ + cos²θ = 1.

Teacher's Note:
1. Add the two and use the identity.

D-4. The value of \( (1 + \tan\theta + \sec\theta)(1 + \cot\theta - \operatorname{cosec}\theta) \) is
(A) 1   (B) \( -1 \)   (C) 2   (D) \( -4 \)
Answer:
Ans. (C) 2
Write in sin and cos: \( \dfrac{(\cos\theta + \sin\theta + 1)(\sin\theta + \cos\theta - 1)}{\sin\theta\cos\theta} = \dfrac{(\sin\theta + \cos\theta)^2 - 1}{\sin\theta\cos\theta} = \dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2 \).
In simple words: Use a common denominator sin θ cos θ.

Teacher's Note:
1. The product is (a + 1)(a − 1) with a = sin θ + cos θ.

D-5. The value of \( 5\tan^2\theta - 5\sec^2\theta \) is
(A) 1   (B) \( -5 \)   (C) 0   (D) 5
Answer:
Ans. (B) \( -5 \)
\( 5(\tan^2\theta - \sec^2\theta) = 5(-1) = -5 \).
In simple words: Take 5 out as a common factor.

Teacher's Note:
1. sec²θ − tan²θ = 1.