CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 03

Question. Prove that: \( \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta \)
Answer: LHS \( = \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)} \)
To obtain \( \cot \theta \) in RHS, we have to convert the numerator of LHS in cosine function and denominator in sin function.
Therefore converting \( \sin^2 \theta = 1 - \cos^2 \theta \), we get
\( = \frac{1 + \cos \theta - (1 - \cos^2 \theta)}{\sin \theta (1 + \cos \theta)} = \frac{1 + \cos \theta - 1 + \cos^2 \theta}{\sin \theta (1 + \cos \theta)} = \frac{\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{\cos \theta (1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \cot \theta = RHS \)

 

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

 

Question. If \( \sec \theta = x + \frac{1}{4x} \), prove that \( \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \).
Answer: Let \( \sec \theta + \tan \theta = \lambda \) ... (i)
We know that, \( \sec^2 \theta - \tan^2 \theta = 1 \)
\( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \implies \lambda (\sec \theta - \tan \theta) = 1 \)
\( \sec \theta - \tan \theta = \frac{1}{\lambda} \) ... (ii)
Adding equations (i) and (ii), we get
\( 2 \sec \theta = \lambda + \frac{1}{\lambda} \)
\( \implies \) \( 2 \left( x + \frac{1}{4x} \right) = \lambda + \frac{1}{\lambda} \)
\( \implies \) \( 2x + \frac{1}{2x} = \lambda + \frac{1}{\lambda} \)
On comparing, we get \( \lambda = 2x \) or \( \lambda = \frac{1}{2x} \)
\( \implies \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \)
Alternative Method:
We have \( \sec \theta = x + \frac{1}{4x} \)
\( \tan^2 \theta = \sec^2 \theta - 1 \)
\( = \left( x + \frac{1}{4x} \right)^2 - 1 \)
\( = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1 \)
\( = x^2 + \frac{1}{16x^2} - \frac{1}{2} = \left( x - \frac{1}{4x} \right)^2 \)
\( \implies \tan \theta = \pm \left( x - \frac{1}{4x} \right) \)
\( \sec \theta + \tan \theta \) is given by
\( x + \frac{1}{4x} + x - \frac{1}{4x} \) or \( x + \frac{1}{4x} - x + \frac{1}{4x} \)
\( = 2x \) or \( \frac{1}{2x} \)

 

Question. If \( \tan (A + B) = 1 \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \), \( 0^\circ < A + B \le 90^\circ \), \( A > B \), then find the values of \( A \) and \( B \). 
Answer: Given \( \tan (A + B) = 1 \).
\( \implies \tan (A + B) = \tan 45^\circ \)
\( \implies A + B = 45^\circ \) — ①
Now taking, \( \tan (A - B) = \frac{1}{\sqrt{3}} \)
\( \implies \tan (A - B) = \tan 30^\circ \)
\( \implies A - B = 30^\circ \) — ②
Adding ① and ②;
\( A + B + A - B = 45^\circ + 30^\circ \)
\( \implies 2A = 75^\circ \implies A = \frac{75^\circ}{2} \implies A = 37.5^\circ \)
\( B = 45^\circ - A \implies B = 45^\circ - 37.5^\circ \implies B = 7.5^\circ \)
\( A = 37.5^\circ, B = 7.5^\circ \) [Topper's Answer 2019]

Long Answer Questions

 

Question. Prove that: \( \frac{\tan^3 \theta}{1 + \tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} = \sec \theta \csc \theta - 2 \sin \theta \cos \theta \) 
Answer: LHS \( = \frac{\tan^3 \theta}{1 + \tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} \)
\( = \frac{\frac{\sin^3 \theta}{\cos^3 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} + \frac{\frac{\cos^3 \theta}{\sin^3 \theta}}{1 + \frac{\cos^2 \theta}{\sin^2 \theta}} \)
\( = \frac{\frac{\sin^3 \theta}{\cos^3 \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} + \frac{\frac{\cos^3 \theta}{\sin^3 \theta}}{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}} \)
\( = \frac{\sin^3 \theta}{\cos^3 \theta} \times \cos^2 \theta + \frac{\cos^3 \theta}{\sin^3 \theta} \times \sin^2 \theta \)
\( = \frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta} \)
\( = \frac{\sin^4 \theta + \cos^4 \theta}{\sin \theta \cos \theta} = \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} \)
\( = \frac{1 - 2 \sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \sec \theta \csc \theta - 2 \sin \theta \cos \theta \)
\( = RHS \) Proved

 

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

 

Question. If \( \tan A = n \tan B \) and \( \sin A = m \sin B \), prove that \( \cos^2 A = \frac{m^2 - 1}{n^2 - 1} \).
Answer: We have to find \( \cos^2 A \) in terms of \( m \) and \( n \). This means that the angle \( B \) is to be eliminated from the given relations.
Now, \( \tan A = n \tan B \)
\( \implies \tan B = \frac{1}{n} \tan A \implies \cot B = \frac{n}{\tan A} \)
and \( \sin A = m \sin B \)
\( \implies \sin B = \frac{1}{m} \sin A \implies \csc B = \frac{m}{\sin A} \)
Substituting the values of \( \cot B \) and \( \csc B \) in \( \csc^2 B - \cot^2 B = 1 \), we get
\( \implies \frac{m^2}{\sin^2 A} - \frac{n^2}{\tan^2 A} = 1 \implies \frac{m^2}{\sin^2 A} - \frac{n^2 \cos^2 A}{\sin^2 A} = 1 \)
\( \implies \frac{m^2 - n^2 \cos^2 A}{\sin^2 A} = 1 \implies m^2 - n^2 \cos^2 A = \sin^2 A \)
\( \implies m^2 - n^2 \cos^2 A = 1 - \cos^2 A \)
\( \implies m^2 - 1 = n^2 \cos^2 A - \cos^2 A \)
\( \implies m^2 - 1 = (n^2 - 1) \cos^2 A \)
\( \implies \frac{m^2 - 1}{n^2 - 1} = \cos^2 A \)

 

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

 

Question. Prove that : \( \frac{(1 + \cot \theta + \tan \theta)(\sin \theta - \cos \theta)}{(\sec^3 \theta - \csc^3 \theta)} = \sin^2 \theta \cos^2 \theta \)
Answer: LHS \( = \frac{(1 + \cot \theta + \tan \theta)(\sin \theta - \cos \theta)}{(\sec^3 \theta - \csc^3 \theta)} \)
\( = \frac{\left( 1 + \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right)(\sin \theta - \cos \theta)}{\left( \frac{1}{\cos^3 \theta} - \frac{1}{\sin^3 \theta} \right)} \)
\( = \frac{\left( \frac{\sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right)(\sin \theta - \cos \theta)}{\left( \frac{\sin^3 \theta - \cos^3 \theta}{\sin^3 \theta \cos^3 \theta} \right)} \)
\( = \frac{\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta}}{\frac{\sin^3 \theta - \cos^3 \theta}{\sin^3 \theta \cos^3 \theta}} = \frac{\sin^3 \theta \cos^3 \theta}{\sin \theta \cos \theta} \)
\( = \sin^2 \theta \cos^2 \theta = RHS \)

 

Question. If \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), show that \( (m^2 - n^2) = 4\sqrt{mn} \). 
Answer: We have, given \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), then
LHS \( = (m^2 - n^2) = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 \)
\( = \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta - \tan^2 \theta - \sin^2 \theta + 2 \tan \theta \sin \theta \)
\( = 4 \tan \theta \sin \theta = 4\sqrt{\tan^2 \theta \sin^2 \theta} \)
\( = 4\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} (1 - \cos^2 \theta)} = 4\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta} \)
\( = 4\sqrt{\tan^2 \theta - \sin^2 \theta} = 4\sqrt{(\tan \theta - \sin \theta)(\tan \theta + \sin \theta)} = 4\sqrt{mn} = RHS \)

 

Question. Prove that: \( \frac{1}{(\csc x + \cot x)} - \frac{1}{\sin x} = \frac{1}{\sin x} - \frac{1}{(\csc x - \cot x)} \).
Answer: In order to show that,
\( \frac{1}{(\csc x + \cot x)} - \frac{1}{\sin x} = \frac{1}{\sin x} - \frac{1}{(\csc x - \cot x)} \) It is sufficient to show
\( \implies \frac{1}{(\csc x + \cot x)} + \frac{1}{(\csc x - \cot x)} = \frac{2}{\sin x} \) ... (i)
Now, LHS of above is
\( = \frac{1}{(\csc x + \cot x)} + \frac{1}{(\csc x - \cot x)} \)
\( = \frac{(\csc x - \cot x) + (\csc x + \cot x)}{(\csc x + \cot x)(\csc x - \cot x)} \)
\( = \frac{2 \csc x}{\csc^2 x - \cot^2 x} \) (\( \because (a + b)(a - b) = a^2 - b^2 \))
\( = \frac{2 \csc x}{1} = \frac{2}{\sin x} = RHS \) of (i)
Hence, \( \frac{1}{(\csc x + \cot x)} + \frac{1}{(\csc x - \cot x)} = \frac{1}{\sin x} + \frac{1}{\sin x} \)
or \( \frac{1}{(\csc x + \cot x)} - \frac{1}{\sin x} = \frac{1}{\sin x} - \frac{1}{(\csc x - \cot x)} \).

 

Question. If \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \) and \( x \sin \theta = y \cos \theta \), prove \( x^2 + y^2 = 1 \). 
Answer: We have, \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \)
\( \implies (x \sin \theta) \sin^2 \theta + (y \cos \theta) \cos^2 \theta = \sin \theta \cos \theta \)
\( \implies x \sin \theta (\sin^2 \theta) + (x \sin \theta) \cos^2 \theta = \sin \theta \cos \theta \) (\( \because x \sin \theta = y \cos \theta \))
\( \implies x \sin \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta \)
\( \implies x \sin \theta = \sin \theta \cos \theta \)
\( \implies x = \cos \theta \)
Now, we have \( x \sin \theta = y \cos \theta \)
\( \implies \cos \theta \sin \theta = y \cos \theta \) (\( \because x = \cos \theta \))
\( \implies y = \sin \theta \)
Hence, \( x^2 + y^2 = \cos^2 \theta + \sin^2 \theta = 1 \).

 

Objective Type Questions: 

 

Question. The value of \( 4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\cos^2 45^\circ - \tan^2 45^\circ) \) is
(a) \( \frac{1}{2} \)
(b) 1
(c) 2
(d) 3
Answer: (c) 2

 

Question. \( \alpha \) is an acute angle \( (\sin \alpha + \cos \alpha) \) is
(a) greater than 1.
(b) less than 1.
(c) equal to 1.
(d) We cannot say any of the options as it depends on the value of \( \alpha \).
Answer: (a) greater than 1.

 

Question. Which of the following option makes the statement below true?
\[ \frac{\frac{1}{\sec x} + \sec x}{\cos^2 x - 1 - \tan^2 x} \]
(a) \( -\csc x \tan x \)
(b) \( -\sec x \tan x \)
(c) \( -\csc x \cot x \)
(d) \( -\sec x \cot x \)
Answer: (c) \( -\csc x \cot x \)

 

Very Short Answer Questions: 

 

Question. If \( \sin A = \frac{3}{4} \), calculate \( \sec A \). 
Answer: \( \frac{4}{\sqrt{7}} \)

 

Question. If \( \sin \theta = \frac{12}{13} \), then find \( \tan \theta \).
Answer: \( \frac{12}{5} \)

 

Question. If \( \csc^2 \theta (1 + \cos \theta)(1 - \cos \theta) = k \), then find the value of \( k \). 
Answer: \( k = 1 \)

 

Question. If \( \tan \alpha = \sqrt{3} \) and \( \tan \beta = \frac{1}{\sqrt{3}} \), then find the value of \( \cot(\alpha + \beta) \).
Answer: 0

 

Question. Evaluate: \( \sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ \) 
Answer: 2

 

Short Answer Questions-I: 

 

Question. What is the maximum value of \( \frac{2}{\csc \theta} \)? Justify your answer.
Answer: 2. Since \( \frac{2}{\csc \theta} = 2 \sin \theta \), and the maximum value of \( \sin \theta \) is 1.

 

Question. If \( \csc \theta = 3x \) and \( \cot \theta = \frac{3}{x} \), then find the value of \( 9\left(x^2 - \frac{1}{x^2}\right) \).
Answer: \( \frac{1}{9} \)

 

Question. What is the value of \( \sin^2 \theta + \frac{1}{1 + \tan^2 \theta} \)?
Answer: 1

 

Short Answer Questions-II:

 

Question. In \( \Delta ABC \), right-angled at \( C \), find \( \cos A \), \( \tan A \) and \( \csc B \) if \( \sin A = \frac{24}{25} \).
Answer: \( \cos A = \frac{7}{25}, \tan A = \frac{24}{7}, \csc B = \frac{25}{7} \)

 

Question. If \( 4 \tan \theta = 3 \), evaluate \( \left(\frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1}\right) \). 
Answer: \( \frac{13}{11} \)

 

Question. Prove that: \( \frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A \)
Answer:
LHS \( = \tan A \left( \frac{1}{1 + \sec A} - \frac{1}{1 - \sec A} \right) \)
\( = \tan A \left( \frac{1 - \sec A - (1 + \sec A)}{1 - \sec^2 A} \right) \)
\( = \tan A \left( \frac{-2 \sec A}{-\tan^2 A} \right) \)
\( = \frac{2 \sec A}{\tan A} = \frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A = RHS \)

 

Question. If \( \cot \theta = \frac{1}{\sqrt{3}} \), show that \( \frac{1 - \cos^2 \theta}{2 - \sin^2 \theta} = \frac{3}{5} \).
Answer:
If \( \cot \theta = \frac{1}{\sqrt{3}} \), then \( \theta = 60^\circ \).
LHS \( = \frac{1 - \cos^2 60^\circ}{2 - \sin^2 60^\circ} = \frac{1 - (1/2)^2}{2 - (\sqrt{3}/2)^2} \)
\( = \frac{1 - 1/4}{2 - 3/4} = \frac{3/4}{5/4} = \frac{3}{5} = RHS \)

 

Question. If \( \sin \theta = \frac{a^2 - b^2}{a^2 + b^2} \), find \( 1 + \tan \theta \cos \theta \).
Answer: \( \frac{2a^2}{a^2 + b^2} \)

 

Question. If \( \sec \theta = \frac{5}{4} \), find the value of \( \frac{\sin \theta - 2 \cos \theta}{\tan \theta - \cot \theta} \).
Answer: \( \frac{12}{7} \)

 

Question. Prove the identity: \( (\csc \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \)
Answer:
LHS \( = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \)
\( = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = RHS \)

 

Question. Prove the identity: \( \cot \theta - \tan \theta = \frac{2 \cos^2 \theta - 1}{\sin \theta \cos \theta} \)
Answer:
LHS \( = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} \)
\( = \frac{\cos^2 \theta - (1 - \cos^2 \theta)}{\sin \theta \cos \theta} = \frac{2 \cos^2 \theta - 1}{\sin \theta \cos \theta} = RHS \)

 

Question. Evaluate: \( \frac{\tan^2 60^\circ + 4 \cos^2 45^\circ + 3 \sec^2 30^\circ}{\csc 30^\circ + \sec 60^\circ - \cot^2 30^\circ} \)
Answer: 9

 

Question. Evaluate: \( \frac{\tan 45^\circ}{\csc 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} - \frac{3}{2} \)
Answer: 1

 

Question. Prove that: \( \frac{(1 + \cot \theta + \tan \theta)(\sin \theta - \cos \theta)}{(\sec^3 \theta - \csc^3 \theta)} = \sin^2 \theta \cos^2 \theta \) 
Answer: Verification by converting into \( \sin \) and \( \cos \).

 

Question. If \( \tan (A + B) = \sqrt{3} \) and \( \tan (A - B) = 0, 0^\circ < A + B \le 90^\circ \), find \( \sin (A + B) \) and \( \cos (A - B) \).
Answer: \( \sin (A + B) = \frac{\sqrt{3}}{2}, \cos (A - B) = 1 \)

 

Long Answer Questions: 

 

Question. Prove that: \( (\sec A - \csc A) (1 + \tan A + \cot A) = \tan A \sec A - \cot A \csc A \)
Answer: Verification by expansion and trigonometric identities.

 

Question. If \( a \sin \theta + b \cos \theta = c \), then prove that \( a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} \).
Answer: Squaring and adding both the relations.

 

Question. Prove that: \( \sin A(1 + \tan A) + \cos A(1 + \cot A) = \sec A + \csc A \)
Answer:
LHS \( = \sin A + \frac{\sin^2 A}{\cos A} + \cos A + \frac{\cos^2 A}{\sin A} \)
\( = \frac{\sin A \cos A + \sin^2 A + \cos^2 A \sin A + \cos^2 A}{\sin A \cos A} \)
Wait, expansion gives: \( = \sin A + \frac{\sin^2 A}{\cos A} + \cos A + \frac{\cos^2 A}{\sin A} \)
\( = (\sin A + \frac{\cos^2 A}{\sin A}) + (\cos A + \frac{\sin^2 A}{\cos A}) \)
\( = \frac{\sin^2 A + \cos^2 A}{\sin A} + \frac{\cos^2 A + \sin^2 A}{\cos A} \)
\( = \frac{1}{\sin A} + \frac{1}{\cos A} = \csc A + \sec A = RHS \)

 

Question. Prove that: \( 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 = 0 \) 
Answer:
Using \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) and \( a^2 + b^2 = (a + b)^2 - 2ab \).
\( 2(1 - 3 \sin^2 \theta \cos^2 \theta) - 3(1 - 2 \sin^2 \theta \cos^2 \theta) + 1 \)
\( = 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1 \)
\( = 3 - 3 = 0 = RHS \)

 

SECTION A

 

Question. Choose and write the correct option in the following questions.
(i) If \( \sin \theta + \cos \theta = \sqrt{2} \cos \theta, (\theta \ne 90^\circ) \), then the value of \( \tan \theta \) is
(a) \( \sqrt{2} - 1 \)
(b) \( \sqrt{2} + 1 \)
(c) \( \sqrt{2} \)
(d) \( -\sqrt{2} \)
Answer: (a) \( \sqrt{2} - 1 \)

 

Question. Given that \( \sin \alpha = \frac{1}{2} \) and \( \cos \beta = \frac{1}{2} \), then the value of \( (\alpha + \beta) \) is
(a) \( 0^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Answer: (d) \( 90^\circ \)

 

Question. \( \frac{1}{\tan \theta + \cot \theta} = \)
(a) \( \cos \theta \sin \theta \)
(b) \( \sec \theta \sin \theta \)
(c) \( \tan \theta \cot \theta \)
(d) \( \sec \theta \csc \theta \)
Answer: (a) \( \cos \theta \sin \theta \)

 

Question. Solve the following questions.
(i) If \( \tan \alpha = \frac{5}{12} \), find the value of \( \sec \alpha \). 
Answer: \( \frac{13}{12} \)

 

Question. (ii) In a right angled triangle if \( \cos \theta = \frac{1}{2}, \sin \theta = \frac{\sqrt{3}}{2} \), what is the value of \( \tan \theta \)?
Answer: \( \sqrt{3} \)

 

SECTION B

 

Question. Find an acute angle \( \theta \), when \( \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \).
Answer: \( 60^\circ \)

 

Question. Evaluate: \( \frac{\cos 60^\circ - \cot 45^\circ + \csc 30^\circ}{\sec 60^\circ + \tan 45^\circ - \sin 30^\circ} \)
Answer: \( -1 \)

 

Question. If \( A = 30^\circ \) and \( B = 30^\circ \), verify that \( \sin (A + B) = \sin A \cos B + \cos A \sin B \).
Answer: Verification: \( \sin(30+30) = \sin 60 = \sqrt{3}/2 \). \( \sin 30 \cos 30 + \cos 30 \sin 30 = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \). LHS = RHS.

 

Question. Solve the following questions.
The altitude \( AD \) of a \( \Delta ABC \), in which \( \angle A \) is an obtuse angle has length 10 cm. If \( BD = 10 \) cm and \( CD = 10\sqrt{3} \) cm, determine \( \angle A \).
Answer: \( 105^\circ \)

 

Trignometry

 

Please click on below link to download CBSE Class 10 Mathematics Trignometry Printable Worksheet Set C