CBSE Class 10 Maths HOTs Introduction to Trigonometry Set 05

Question. If \( x = 3 \sec^2 \theta - 1 \), \( y = \tan^2 \theta - 2 \) then \( x - 3y \) is equal to
(a) 3
(b) 4
(c) 8
(d) 5
Answer: (c) 8

Question. If \( \sec \theta - \tan \theta = \frac{1}{3} \), the value of \( (\sec \theta + \tan \theta) \) is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3

Question. The value of \( \cos 60^\circ \sin 30^\circ + \sin 60^\circ \cos 30^\circ \) is :
(a) \( \frac{1}{4} \)
(b) \( \frac{3}{4} \)
(c) 1
(d) \( \frac{2}{4} \)
Answer: (c) 1

Question. If \( \cot \theta = \frac{7}{8} \), the value of \( \frac{(1 + \cos \theta)(1 - \cos \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \) is :
(a) \( \frac{49}{64} \)
(b) \( \frac{8}{7} \)
(c) \( \frac{64}{49} \)
(d) \( \frac{7}{8} \)
Answer: (a) \( \frac{49}{64} \)

Question. If \( \cos (20 + \theta) = \sin 30^\circ \), then the value of \( \theta \) is :
(a) \( 20^\circ \)
(b) \( 50^\circ \)
(c) \( 30^\circ \)
(d) \( 40^\circ \)
Answer: (d) \( 40^\circ \)

Question. The value of \( \sin \theta \cos(90^\circ - \theta) + \cos \theta \sin(90^\circ - \theta) \) is:
(a) 1
(b) 0
(c) 2
(d) -1
Answer: (a) 1

Question. If \( \csc \theta = \frac{3}{2} \), then \( 2(\csc^2 \theta + \cot^2 \theta) \) is:
(a) 3
(b) 7
(c) 9
(d) 5
Answer: (b) 7

Question. If \( \sin \theta + \sin^2 \theta = 1 \), the value of \( (\cos^2 \theta + \cos^4 \theta) \) is :
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (c) 1

Question. The value of \( \sin^2 30^\circ + \cos^2 45^\circ + \cos^2 30^\circ \) is :
(a) \( \frac{1}{2} \)
(b) \( \frac{\sqrt{3}}{2} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{2}{3} \)
Answer: (c) \( \frac{3}{2} \)

Question. If \( \sin 2\theta = \cos (\theta - 6^\circ) \) where \( 2\theta \) and \( (\theta - 6^\circ) \) are both acute angles then the value of \( \theta \) is :
(a) \( 16^\circ \)
(b) \( 32^\circ \)
(c) \( 48^\circ \)
(d) \( 45^\circ \)
Answer: (b) \( 32^\circ \)

Question. Given that \( \cos \theta = \frac{1}{2} \), the value of \( \frac{2 \sec \theta}{1 + \tan^2 \theta} \) is :
(a) 1
(b) 2
(c) \( \frac{1}{2} \)
(d) 0
Answer: (a) 1

Question. If \( \sec A = \csc B = \frac{13}{12} \) then \( A + B \) is equal to:
(a) Zero
(b) \( >90^\circ \)
(c) \( 90^\circ \)
(d) \( 40 \)
Answer: (c) \( 90^\circ \)

Question. If \( \cot A = \frac{12}{5} \), then the value of \( (\sin A + \cos A) \times \csc A \) is :
(a) \( \frac{13}{5} \)
(b) \( \frac{17}{5} \)
(c) \( \frac{14}{5} \)
(d) 1
Answer: (b) \( \frac{17}{5} \)

Question. \( 9 \sec^2 \theta - 9 \tan^2 \theta \) is equal to :
(a) 1
(b) -1
(c) 9
(d) -9
Answer: (c) 9

Question. In \( \triangle ABC \) right angled at B, \( \tan A = 1 \), the value of \( 2 \sin A \cos A \) is :
(a) -1
(b) 2
(c) 3
(d) 1
Answer: (d) 1

Question. If \( \sqrt{2} \sin (60 - \alpha) = 1 \), then the value of \( \alpha \) is :
(a) \( 45^\circ \)
(b) \( 15^\circ \)
(c) \( 60^\circ \)
(d) \( 30^\circ \)
Answer: (b) \( 15^\circ \)

Question. If \( \tan (A - B) = \frac{1}{\sqrt{3}} \) and \( \sin A = \frac{1}{\sqrt{2}} \), then the value of \( B \) is :
(a) \( 45^\circ \)
(b) \( 60^\circ \)
(c) \( 0^\circ \)
(d) \( 15^\circ \)
Answer: (d) \( 15^\circ \)

Very Short Answer Type Questions 

Question. Find the value of the expression \( \frac{\cos 30^\circ + \sin 60^\circ}{1 + \cos 60^\circ + \sin 30^\circ} \)
Answer: \( \pm \frac{\sqrt{3}}{2} \)

Question. If \( \angle A \) and \( \angle B \) are acute angles such that \( \cos A = \cos B \). Show that \( \angle A = \angle B \).
Answer: By definition of trigonometric ratios in a right-angled triangle, if \( \cos A = \cos B \), then the adjacent side over hypotenuse for both angles must be proportional, implying the angles are equal for acute cases.

Question. Evaluate \( \frac{\csc^2 (90^\circ - \theta) - \tan^2 \theta}{4(\cos^2 48^\circ + \cos^2 42^\circ)} - \frac{2 \tan^2 30^\circ \sec^2 52^\circ \sin^2 38^\circ}{\csc^2 70^\circ - \tan^2 20^\circ} \)
Answer: The expression simplifies using trigonometric identities \( \csc(90^\circ-\theta) = \sec \theta \), \( \sec^2 \theta - \tan^2 \theta = 1 \), and complementary angle relations.

Question. Prove that \( 1 + \frac{\cot^2 \theta}{1 + \csc \theta} = \frac{1}{\sin \theta} \).
Answer: LHS \( = 1 + \frac{\csc^2 \theta - 1}{1 + \csc \theta} = 1 + \frac{(\csc \theta - 1)(\csc \theta + 1)}{\csc \theta + 1} = 1 + \csc \theta - 1 = \csc \theta = \frac{1}{\sin \theta} = \) RHS.

Question. If \( \sqrt{3} \tan \theta = 3 \sin \theta \), then prove that \( \sin^2 \theta - \cos^2 \theta = \frac{1}{3} \).
Answer: \( \sqrt{3} \frac{\sin \theta}{\cos \theta} = 3 \sin \theta
\( \implies \) \( \cos \theta = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \).
Now, \( \sin^2 \theta - \cos^2 \theta = (1 - \cos^2 \theta) - \cos^2 \theta = 1 - 2\cos^2 \theta = 1 - 2\left(\frac{1}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3} \).

Question. If \( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \), then prove that \( \sec \theta + \csc \theta = 2 + \frac{2}{\sqrt{3}} \).
Answer: Divide by \( \cos^2 \theta \) or use \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \tan \theta = \frac{1}{\sqrt{3}} \), giving \( \theta = 30^\circ \). Substituting \( \theta \) in \( \sec \theta + \csc \theta \) yields the result.

Question. Given that \( \sin (A + B) = \sin A \cos B + \cos A \sin B \), find the value of \( \sin 75^\circ \)
Answer: \( \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Prove that \( \frac{\sin \theta}{\sin (90^\circ - \theta)} + \frac{\cos \theta}{\cos (90^\circ - \theta)} = \sec \theta \cdot \csc \theta \).
Answer: LHS \( = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \sec \theta \cdot \csc \theta = \) RHS.

Question. If \( \sin (A - B) = \frac{1}{2} \), \( \cos (A + B) = \frac{1}{2} \) and \( 0 < A + B < 90^\circ \) and \( A > B \) then find the values of A and B.
Answer: \( A = 45^\circ, B = 15^\circ \)

Question. If \( \sin \theta + \cos \theta = m \) and \( \sec \theta + \csc \theta = n \), then prove that \( n (m^2 - 1) = 2m \).
Answer: \( m^2 = (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta \).
\( m^2 - 1 = 2 \sin \theta \cos \theta \).
\( n = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{m}{\sin \theta \cos \theta} \).
Now, \( n(m^2 - 1) = \frac{m}{\sin \theta \cos \theta} \times 2 \sin \theta \cos \theta = 2m \).

 VALUE BASED QUESTIONS

Question. How many trigonometrical ratios are there in all and why? Explain the values of their existence.
Answer: In a right angled triangle we have three sides i.e. Perpendicular, hypotenuse and base. To find the ratio we need two sides at a time hence only 6 such possibilities are there, which are ratio of perpendicular to Hypotenuse, Base to Hypotenuse, Perpendicular : base, Base : perpendicular, Hypotenuse to base, Hypotenuse to perpendicular.
Values:

• We should keen to learn and grab the new dimension of any topic.
• We should try to interrelate the relations each other also like, \( \sin \theta = \frac{1}{\text{cosec } \theta} \), \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) etc.

Question. What are basic trigonometric identities. Also explain the values of identities?
Answer: \( H^2 = P^2 + B^2 \) (by pythagoras theorem) so if we divide both sides by \( H^2, B^2 \) and \( P^2 \) respectively then we get \( \left( \frac{P}{H} \right)^2 + \left( \frac{B}{H} \right)^2 = \left( \frac{H}{H} \right)^2 \) i.e. \( \sin^2 \theta + \cos^2 \theta = 1 \). Similarly \( 1 + \tan^2 \theta = \sec^2 \theta \) and \( 1 + \cot^2 \theta = \text{cosec}^2 \theta \). So there are three basic trigonometric identities.
Values:

• Identities are those which are always correct for any value of variable so same way in life we should be always truthful in any condition of variables.

Question. Prove that \( \sin 45^\circ = \frac{1}{\sqrt{2}} \). What value will you conclude?
Answer: In a right triangle with one angle \( 45^\circ \) it is always isosceles i.e. angles will be \( 90^\circ, 45^\circ \) and \( 45^\circ \). Let length of \( AB = x \) units so \( BC = x \) units and by pythagoras theorem \( AC = x \sqrt{2} \) units hence \( \sin 45^\circ = \frac{AB}{AC} \text{ or } \frac{BC}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \).
Values:

• To solve problems of trigonometry we take help of topic geometry so same way we need to help each other in society to know new things.

Question. Prove that \( \sin^2 31^\circ + \sin^2 59^\circ = 1 \). Find the value which can be extracted.
Answer: We have \( \sin^2 31^\circ + \sin^2 59^\circ = \sin^2 31^\circ + \sin^2 (90^\circ - 31^\circ) = \sin^2 31^\circ + \cos^2 31^\circ \text{ [} \because \sin (90^\circ - \theta) = \cos \theta \text{]} \)
\( \implies \) \( \sin^2 31^\circ + \cos^2 31^\circ = 1 \) [using identity \( \sin^2 \theta + \cos^2 \theta = 1 \)]
Values:

• Values as to complete any identity we need to complement in our day to day life for making a healthy relations.