CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 04

Multiple Choice Questions 

Question. \(\tan A =\)
(a) \(\frac{\cos A}{\sqrt{1 - \cos^2 A}}\)
(b) \(\frac{\sec A}{\sqrt{1 - \sec^2 A}}\)
(c) \(\frac{\sin A}{\sqrt{1 - \sin^2 A}}\)
(d) \(\frac{1}{\sqrt{1 - \sin^2 A}}\)
Answer: (c) \(\frac{\sin A}{\sqrt{1 - \sin^2 A}}\)

 

Question. If \(\text{cosec } A - \cot A = \frac{4}{5}\), then \(\text{cosec } A =\)
(a) \(\frac{47}{40}\)
(b) \(\frac{59}{40}\)
(c) \(\frac{51}{40}\)
(d) \(\frac{41}{40}\)
Answer: (d) \(\frac{41}{40}\)

 

Question. If \(\sin x + \text{cosec } x = 2\), then \(\sin^{19} x + \text{cosec}^{20} x =\)
(a) \(2^{19}\)
(b) \(2^{20}\)
(c) 2
(d) \(2^{39}\)
Answer: (c) 2

 

Question. If \(\tan A + \cot A = 4\), then \(\tan^4 A + \cot^4 A\) is equal to
(a) 190
(b) 184
(c) 194
(d) 180
Answer: (c) 194

 

Question. If \(\tan A = \frac{5}{12}\), find the value of \((\sin A + \cos A) \cdot \sec A\).
(a) \(\frac{17}{12}\)
(b) \(\frac{19}{12}\)
(c) \(\frac{21}{12}\)
(d) \(\frac{11}{12}\)
Answer: (a) \(\frac{17}{12}\)

 

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

 

Question. If \(\sin \theta = \frac{1}{3}\), then find the value of \((2 \cot^2 \theta + 2)\)
(a) 9
(b) 12
(c) 15
(d) 18
Answer: (d) 18

 

Question. If \(3x = \text{cosec } \theta\) and \(\frac{3}{x} = \cot \theta\), find the value of \(3\left(x^2 - \frac{1}{x^2}\right)\).
(a) \(\frac{1}{9}\)
(b) 9
(c) \(\frac{1}{3}\)
(d) 3
Answer: (c) \(\frac{1}{3}\)

 

Question. The value of \( \sin^2 30^\circ - \cos^2 30^\circ \) is
(a) \( -\frac{1}{2} \)
(b) \( \frac{\sqrt{3}}{2} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{2}{3} \)
Answer: (a) \( -\frac{1}{2} \)
\( \sin^2 30^\circ - \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \) \( \frac{1}{4} - \frac{3}{4} = \frac{-2}{4} = \frac{-1}{2} \).

 

Question. If \( 3 \cot \theta = 2 \), then the value of \( \tan \theta \) is
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{2} \)
(c) \( \frac{3}{\sqrt{13}} \)
(d) \( \frac{2}{\sqrt{13}} \)
Answer: (b) \( \frac{3}{2} \)
\( 3 \cot \theta = 2 \)
\( \implies \) \( \cot \theta = \frac{2}{3} \)
\( \implies \) \( \tan \theta = \frac{3}{2} \).

 

Question. If \( \sin A = \cos A \), \( 0^\circ < A < 90^\circ \), then \( A \) is equal to
(a) \( 60^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 90^\circ \)
Answer: (b) \( 45^\circ \)
\( \sin A = \cos A \)
\( \implies \) \( \frac{\sin A}{\cos A} = 1 \)
\( \implies \) \( \tan A = 1 \)
\( \implies \) \( A = 45^\circ \).

 

Question. Evaluate: \( \sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ \) 
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
\( \left(\frac{\sqrt{3}}{2}\right)^2 + 2(1) - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \) \( \frac{3}{4} + 2 - \frac{3}{4} = 2 \).

 

Question. If \( \tan \theta = \frac{1}{\sqrt{3}} \), then evaluate \( \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} \)
(a) 1
(b) 2
(c) \( \frac{1}{2} \)
(d) -1
Answer: (c) \( \frac{1}{2} \)
\( \tan \theta = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \theta = 30^\circ \)
\( \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} = \frac{\text{cosec}^2 30^\circ - \sec^2 30^\circ}{\text{cosec}^2 30^\circ + \sec^2 30^\circ} \)
\( \implies \) \( \frac{(2)^2 - \left(\frac{2}{\sqrt{3}}\right)^2}{(2)^2 + \left(\frac{2}{\sqrt{3}}\right)^2} = \frac{4 - \frac{4}{3}}{4 + \frac{4}{3}} = \frac{1}{2} \).

 

Question. If \( \sin(A - B) = \frac{1}{2} \), \( \cos(A + B) = \frac{1}{2} \), find \( A \) and \( B \).
(a) \( 30^\circ, 45^\circ \)
(b) \( 45^\circ, 30^\circ \)
(c) \( 45^\circ, 15^\circ \)
(d) \( 60^\circ, 15^\circ \)
Answer: (c) \( 45^\circ, 15^\circ \)
\( \sin(A - B) = \frac{1}{2} \)
\( \implies \) \( A - B = 30^\circ \) ...(i)
and \( \cos(A + B) = \frac{1}{2} \)
\( \implies \) \( A + B = 60^\circ \) ...(ii)
Solving equations (i) and (ii), we get \( A = 45^\circ \) and \( B = 15^\circ \).

 

Very Short Answer Type Questions 

 

Question. If \( \sqrt{3} \sin \theta - \cos \theta = 0 \) and \( 0^\circ < \theta < 90^\circ \), find the value of \( \theta \).
Answer: \( \sqrt{3} \sin \theta - \cos \theta = 0 \)
\( \implies \) \( \sqrt{3} \sin \theta = \cos \theta \)
\( \implies \) \( \sqrt{3} = \frac{\cos \theta}{\sin \theta} \)
\( \implies \) \( \cot \theta = \sqrt{3} \implies \theta = 30^\circ \).

 

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

 

Question. Evaluate : \( 4 \cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ \).
Answer: \( 4(\cot 45^\circ)^2 - (\sec 60^\circ)^2 + (\sin 60^\circ)^2 + (\cos 90^\circ)^2 \)
\( \implies \) \( 4 \times (1)^2 - (2)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + 0 = 4 - 4 + \frac{3}{4} + 0 = \frac{3}{4} \).

 

Question. Find the value of \( \theta \) if \( 2 \sin 2\theta = \sqrt{3} \)
Answer: \( 2 \sin 2\theta = \sqrt{3} \)
\( \implies \) \( \sin 2\theta = \frac{\sqrt{3}}{2} \)
Also, \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \implies \) \( \sin 2\theta = \sin 60^\circ \)
\( \implies \) \( 2\theta = 60^\circ \implies \theta = 30^\circ \).

 

Question. Find the value of \( x \) if \( \tan 3x = \sin 45^\circ \cdot \cos 45^\circ + \sin 30^\circ \).
Answer: \( \tan 3x = \sin 45^\circ \cdot \cos 45^\circ + \sin 30^\circ \)
\( \implies \tan 3x = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \)
\( \implies \tan 3x = \frac{1}{2} + \frac{1}{2} = 1 \)
\( \implies \tan 3x = \tan 45^\circ \)
\( \implies 3x = 45^\circ \implies x = 15^\circ \).

 

Short Answer Type Question

 

Question. If \( \sin(A + 2B) = \frac{\sqrt{3}}{2} \) and \( \cos(A + 4B) = 0 \), \( A > B \), and \( A + 4B \leq 90^\circ \), then find \( A \) and \( B \). 
Answer: \( \sin(A + 2B) = \frac{\sqrt{3}}{2} \)
\( \implies A + 2B = 60^\circ \) ...(i)
And \( \cos(A + 4B) = 0 \implies \cos(A + 4B) = \cos 90^\circ \)
\( \implies A + 4B = 90^\circ \) ...(ii)
Subtracting (i) from (ii), we get:
\( 2B = 30^\circ \implies B = 15^\circ \).
Putting \( B = 15^\circ \) in eq. (i), we have:
\( A + 2(15^\circ) = 60^\circ \implies A + 30^\circ = 60^\circ \implies A = 30^\circ \).

 

Long Answer Type Questions 

 

Question. Evaluate: \( \frac{2 \cos^2 90^\circ + 4 \cos^2 45^\circ + \tan^2 60^\circ + 3 \text{cosec}^2 60^\circ + 1}{3 \sec 60^\circ - \frac{7}{2} \sec^2 45^\circ + 2 \text{cosec } 30^\circ - 1} \) 
Answer: \( \frac{2(0)^2 + 4(\frac{1}{\sqrt{2}})^2 + (\sqrt{3})^2 + 3(\frac{2}{\sqrt{3}})^2 + 1}{3(2) - \frac{7}{2}(\sqrt{2})^2 + 2(2) - 1} \)
\( \implies \frac{0 + 4(\frac{1}{2}) + 3 + 3(\frac{4}{3}) + 1}{6 - \frac{7}{2}(2) + 4 - 1} = \frac{2 + 3 + 4 + 1}{6 - 7 + 4 - 1} = \frac{10}{2} = 5 \).

 

Question. If \( \sin \theta \) and \( \sec \theta \) are the roots of the equation \( \sqrt{3}x^2 + kx + 3 = 0 \), then find the value of \( k \).
Answer: \( \sin \theta + \sec \theta = \frac{-k}{\sqrt{3}} \) (Sum of the roots)
And \( \sin \theta \cdot \sec \theta = \frac{3}{\sqrt{3}} \) (Product of the roots)
\( \implies \frac{\sin \theta}{\cos \theta} = \sqrt{3} \implies \tan \theta = \sqrt{3} \implies \theta = 60^\circ \).
Putting \( \theta = 60^\circ \) in the sum equation:
\( \sin 60^\circ + \sec 60^\circ = \frac{-k}{\sqrt{3}} \)
\( \implies \frac{\sqrt{3}}{2} + 2 = \frac{-k}{\sqrt{3}} \)
\( \implies \frac{\sqrt{3} + 4}{2} = \frac{-k}{\sqrt{3}} \)
\( \implies k = -\left( \frac{3 + 4\sqrt{3}}{2} \right) \).

 

Question. Ratios of sides of a right triangle with respect to its acute angles are known as
(a) trigonometric identities
(b) trigonometry
(c) trigonometric ratios of the angles
(d) None of the options
Answer: (c) trigonometric ratios of the angles

 

Question. If \( \tan \theta = \frac{a}{b} \), then the value of \( \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} \) is
(a) \( \frac{a^2 - b^2}{a^2 + b^2} \)
(b) \( \frac{a^2 - b^2}{a^2 + b^2} \)
(c) \( \frac{a}{a^2 + b^2} \)
(d) \( \frac{a^2 + b^2}{b} \)
Answer: (b) \( \frac{a^2 - b^2}{a^2 + b^2} \)

 

Question. Match the Columns:
1. \( \frac{\text{Side opposite to angle } \theta}{\text{Hypotenuse}} \) (A) \( \tan \theta \)
2. \( \frac{\text{Side adjacent to angle } \theta}{\text{Hypotenuse}} \) (B) \( \sin \theta \)
3. \( \frac{\text{Side opposite to angle } \theta}{\text{Side adjacent to angle } \theta} \) (C) \( \cos \theta \)
                                               (D) \( \sec \theta \)
(a) 1 – A, 2 – C, 3 – B
(b) 1 – B, 2 – C, 3 – A
(c) 1 – B, 2 – C, 3 – D
(d) 1 – D, 2 – B, 3 – A
Answer: (b) 1 – B, 2 – C, 3 – A

 

Question. Consider the triangle shown below. What are the values of \( \tan \theta \), \( \csc \theta \) and \( \sec \theta \)?
(a) \( \tan \theta = \frac{8}{15}, \csc \theta = \frac{17}{15}, \sec \theta = \frac{17}{8} \)
(b) \( \tan \theta = \frac{8}{15}, \csc \theta = \frac{17}{8}, \sec \theta = \frac{17}{15} \)
(c) \( \tan \theta = \frac{17}{15}, \csc \theta = \frac{8}{15}, \sec \theta = \frac{8}{17} \)
(d) \( \tan \theta = \frac{8}{15}, \csc \theta = \frac{17}{15}, \sec \theta = \frac{17}{8} \)
Answer: (b) \( \tan \theta = \frac{8}{15}, \csc \theta = \frac{17}{8}, \sec \theta = \frac{17}{15} \)

 

Question. In \( \Delta ABC \), right angled at B, AB = 5 cm and \( \sin C = \frac{1}{2} \). The length of side AC is
(a) 12
(b) 2
(c) 6
(d) 10
Answer: (d) 10

 

Question. If \( \sec \theta = \frac{25}{7} \), find the value of \( \tan \theta \).
(a) 24/7
(b) 7/24
(c) 25/7
(d) 25/24
Answer: (a) 24/7

 

Question. In \( \Delta ABC \), right angled at B, if AB = 12 cm and BC = 5 cm, find \( \sin A \)
(a) 13/5
(b) 5/13
(c) 12/13
(d) 5/12
Answer: (b) 5/13

 

Question. If \( 3 \tan \theta = 4 \), find the value of \( \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} \).
Answer: \( 3 \tan \theta = 4 \)
\( \implies \) \( \tan \theta = \frac{4}{3} \)
Now given expression is \( \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} \)
Dividing numerator and denominator by \( \cos \theta \), we get
\( \frac{5 \tan \theta - 3}{5 \tan \theta + 2} \)
Putting \( \tan \theta = \frac{4}{3} \), we get,
\( \frac{5 \times \frac{4}{3} - 3}{5 \times \frac{4}{3} + 2} = \frac{\frac{20}{3} - 3}{\frac{20}{3} + 2} = \frac{11/3}{26/3} = \frac{11}{26} \)

 

Question. If \( \sin \theta + \csc \theta = 3 \), then find the value of \( \frac{\sin^4 \theta + 1}{\sin^2 \theta} \).
Answer: Given equation is
\( \sin \theta + \csc \theta = 3 \)
\( \implies \) \( \sin \theta + \frac{1}{\sin \theta} = 3 \)
\( \implies \) \( \frac{\sin^2 \theta + 1}{\sin \theta} = 3 \)
Squaring both sides we get
\( \left(\frac{\sin^2 \theta + 1}{\sin \theta}\right)^2 = (3)^2 \)
\( \implies \) \( \frac{\sin^4 \theta + 1 + 2 \sin^2 \theta}{\sin^2 \theta} = 9 \)
\( \implies \) \( \frac{\sin^4 \theta + 1}{\sin^2 \theta} + \frac{2 \sin^2 \theta}{\sin^2 \theta} = 9 \)
\( \implies \) \( \frac{\sin^4 \theta + 1}{\sin^2 \theta} + 2 = 9 \)
\( \therefore \) \( \frac{\sin^4 \theta + 1}{\sin^2 \theta} = 7 \)

 

PRACTICE QUESTIONS

Question. Match the Columns:
1. \( \frac{\text{Hypotenuse}}{\text{Side adjacent to angle } \theta} \)  (A) \( \csc \theta \)
2. \( \frac{\text{Side adjacent to angle } \theta}{\text{Side opposite to angle } \theta} \)  (B) \( \sec \theta \)
3. \( \frac{\text{Hypotenuse}}{\text{Side opposite to angle } \theta} \)  (C) \( \tan \theta \)
                                         (D) \( \cot \theta \)
                                        (E) \( \sin \theta \)
(a) 1 – A, 2 – C, 3 – B
(b) 1 – C, 2 – A, 3 – D
(c) 1 – B, 2 – A, 3 – E
(d) 1 – B, 2 – D, 3 – A
Answer: (d) 1 – B, 2 – D, 3 – A

 

Question. Reciprocal of \( \csc A \) is
(a) \( \sec A \)
(b) \( \cos A \)
(c) \( \sin A \)
(d) \( \cot A \)
Answer: (c) \( \sin A \)

 

Question. If \( \tan \theta = \frac{4}{3} \), then find the value of \( \left[\frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta}\right] \). 
Answer: 7

 

Question. If \( \cot A = \frac{5}{12} \), then find the value of \( (\sin A + \cos A) \csc A \).
Answer: 17/12

 

Question. If \( \sin \theta = x \) and \( \sec \theta = y \), then find the value of \( \cot \theta \).
Answer: \( \frac{1}{xy} \)

 

Question. If \( \tan \theta + \frac{1}{\tan \theta} = 2 \), find the value of \( \tan^2 \theta + \frac{1}{\tan^2 \theta} \).
Answer: 2

 

Question. If \( \tan \theta = \frac{p}{q} \), show that \( \frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta} = \frac{p^2 - q^2}{p^2 + q^2} \).
Answer: LHS \( = \frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta} \)
Dividing by \( \cos \theta \):
\( \implies \frac{p \tan \theta - q}{p \tan \theta + q} \)
Substituting \( \tan \theta = \frac{p}{q} \):
\( \implies \frac{p(p/q) - q}{p(p/q) + q} \)
\( \implies \frac{p^2/q - q}{p^2/q + q} = \frac{p^2 - q^2}{p^2 + q^2} = \) RHS. Hence Proved.

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

 

Question. Assertion (A) : \( \sin^2 67^\circ + \cos^2 67^\circ = 1 \).
Reason (R) : For any value of \( \theta \), \( \sin^2 \theta + \cos^2 \theta = 1 \)
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution : \( \sin^2 \theta + \cos^2 \theta = 1 \)

\( \implies \sin^2 67^\circ + \cos^2 67^\circ = 1 \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

 

Question. Assertion (A) : If \( \cos A + \cos^2 A = 1 \) then \( \sin^2 A + \sin^4 A = 2 \)
Reason (R) : \( 1 - \sin^2 A = \cos^2 A \), for any value of \( A \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.
Solution : \( \cos A + \cos^2 A = 1 \)

\( \implies \cos A = 1 - \cos^2 A = \sin^2 A \)
\( \therefore \sin^2 A + \sin^4 A = \cos A + \cos^2 A = 1 \)

\( \implies \sin^2 A + \sin^4 A = 1 \)
So, A is false but R is true.
Hence, option (d) is correct.

 

Question. Assertion (A) : In a right angled triangle, if \( \tan \theta = \frac{3}{4} \) then \( \sin \theta = \frac{3}{5} \).
Reason (R) : \( \sin 60^\circ = \frac{1}{2} \)
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.
Solution : Let \( \Delta ABC \) be a right angled triangle and \( \angle C = \theta \).
It is given that \( \tan \theta = \frac{3}{4} = \frac{AB}{BC} \)
Let \( AB = 3K, BC = 4K \)
\( \therefore AC = \sqrt{AB^2 + BC^2} = \sqrt{9K^2 + 16K^2} = 5K \)
\( \therefore \sin \theta = \frac{AB}{AC} = \frac{3K}{5K} = \frac{3}{5} \)
Thus, Assertion (A) is true.
But \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), \( \therefore \) Reason (R) is false.
Hence, option (c) is correct.

 

Question. Assertion (A) : The value of \( \sin \theta = \frac{4}{3} \) is not possible.
Reason (R) : Hypotenuse is the longest side in any right angled triangle.
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution : \( \sin \theta = \frac{P}{H} = \frac{4k}{3k} \)
Here, perpendicular is greater than the hypotenuse which is not possible as in any right triangle hypotenuse is the longest side.
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

 

 

Trignometry

 

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