CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 08

Question. In \(\Delta ABC\), right angled at B, \( AC = 13 \) cm, \( AB = 5 \) cm, then \(\sin A\) is equal to
(a) \( \frac{5}{13} \)
(b) \( \frac{5}{12} \)
(c) \( \frac{12}{13} \)
(d) \( \frac{13}{12} \)
Answer: (c) \( \frac{12}{13} \)

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

Question. The maximum value of \( \frac{1}{\text{cosec } \theta} \) is
(a) 0
(b) 1
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \frac{1}{\sqrt{2}} \)
Answer: (b) 1

Question. The value of \( \cos 0^\circ \cdot \cos 1^\circ \cdot \cos 2^\circ \dots \cos 90^\circ \) is equal to
(a) 1
(b) 2
(c) 3
(d) None of the options
Answer: (d) None of the options

Question. If \( x = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) \), then \( x = \)
(a) 0
(b) 1
(c) – 1
(d) \(\pm 1\)
Answer: (d) \(\pm 1\)

Question. The value of \( 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 = \)
(a) 1
(b) 2
(c) 4
(d) 0
Answer: (d) 0

Question. In \(\Delta ABC\) right angled at B, \(\sin A = \frac{7}{25}\), then the value of \(\cos C\) is
(a) \( \frac{7}{25} \)
(b) \( \frac{24}{25} \)
(c) \( \frac{7}{24} \)
(d) \( \frac{24}{7} \)
Answer: (a) \( \frac{7}{25} \)

Question. Given that \( \sec \theta = \sqrt{2} \), the value of \( \frac{1 + \tan \theta}{\sin \theta} \) is
(a) \( 2\sqrt{2} \)
(b) \( \sqrt{2} \)
(c) \( 3\sqrt{2} \)
(d) 2
Answer: (a) \( 2\sqrt{2} \)

Question. If \( \sec \theta + \tan \theta = p \), then \(\tan \theta\) is
(a) \( \frac{p^2 + 1}{2p} \)
(b) \( \frac{p^2 - 1}{2p} \)
(c) \( \frac{p^2 - 1}{p^2 + 1} \)
(d) \( \frac{p^2 + 1}{p^2 - 1} \)
Answer: (b) \( \frac{p^2 - 1}{2p} \)

Question. Name the study of relationship between the sides and angles of a triangle.
Answer: Trigonometry

Question. In a right triangle, the hypotenuse is 2 times as long as its one side. Find one of the acute angle.
Answer: Let one side be \( a \) and hypotenuse be \( 2a \).
Then, \( \sin \theta = \frac{a}{2a} = \frac{1}{2} \)
\( \implies \theta = 30^\circ \).
The other acute angle would be \( 90^\circ - 30^\circ = 60^\circ \).

Question. If \(\text{cosec } \theta - \cot \theta = \frac{1}{3}\), find the value of \((\text{cosec } \theta + \cot \theta)\).
Answer: We know that \( \text{cosec}^2 \theta - \cot^2 \theta = 1 \)
\( \implies (\text{cosec } \theta - \cot \theta)(\text{cosec } \theta + \cot \theta) = 1 \)
\( \implies \frac{1}{3}(\text{cosec } \theta + \cot \theta) = 1 \)
\( \implies (\text{cosec } \theta + \cot \theta) = 3 \)

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

Question. If \( \cot \theta = \frac{15}{8} \), then evaluate \( \frac{(2 + 2\sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(2 - 2\cos \theta)} \).
Answer: \( \frac{2(1 + \sin \theta)(1 - \sin \theta)}{2(1 + \cos \theta)(1 - \cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \)
\( = \left( \frac{15}{8} \right)^2 = \frac{225}{64} \)

Question. Find the value of \(\text{cosec } 30^\circ\) geometrically.
Answer: Consider an equilateral triangle \( ABC \) with side \( 2a \). Draw an altitude \( AD \). In \(\Delta ABD\), \( AB = 2a \), \( BD = a \), and \( \angle BAD = 30^\circ \).
\( \text{cosec } 30^\circ = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{AB}{BD} = \frac{2a}{a} = 2 \).

Question. If \( \sqrt{3} \sin \theta - \cos \theta = 0 \) and \( 0^\circ < \theta < 90^\circ \), find the value of \(\theta\). [CBSE 2012]
Answer: \( \sqrt{3} \sin \theta = \cos \theta \)
\( \implies \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \)
\( \implies \tan \theta = \frac{1}{\sqrt{3}} \)
\( \implies \theta = 30^\circ \)

Question. Show that \( (1 + \cot \theta - \text{cosec } \theta) (1 + \tan \theta + \sec \theta) = 2 \).
Answer: Converting to sin and cos:
\( \left( 1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \right) \left( 1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} \right) \)
\( = \left( \frac{\sin \theta + \cos \theta - 1}{\sin \theta} \right) \left( \frac{\cos \theta + \sin \theta + 1}{\cos \theta} \right) \)
\( = \frac{(\sin \theta + \cos \theta)^2 - 1^2}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \)
\( = \frac{1 + 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2 \)

Question. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), show that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).
Answer: \( \sin \theta = \sqrt{2} \cos \theta - \cos \theta = (\sqrt{2} - 1) \cos \theta \)
Multiplying both sides by \( (\sqrt{2} + 1) \):
\( (\sqrt{2} + 1) \sin \theta = (\sqrt{2} + 1)(\sqrt{2} - 1) \cos \theta \)
\( \sqrt{2} \sin \theta + \sin \theta = (2 - 1) \cos \theta = \cos \theta \)
\( \implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta \)

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

Question. \(ABC\) is a right triangle, right angled at C. If A = 30° and AB = 40 units, find the remaining two sides of \(\Delta ABC\).
Answer: \( \sin 30^\circ = \frac{BC}{AB} \implies \frac{1}{2} = \frac{BC}{40} \implies BC = 20 \) units.
\( \cos 30^\circ = \frac{AC}{AB} \implies \frac{\sqrt{3}}{2} = \frac{AC}{40} \implies AC = 20\sqrt{3} \) units.

Question. Find the value of other trigonometric ratios, given that \(\cos \theta = \frac{2m}{m^2 + 1}\).
Answer: Let Base \(= 2m\), Hypotenuse \(= m^2 + 1\).
Perpendicular \(= \sqrt{(m^2 + 1)^2 - (2m)^2} = \sqrt{m^4 + 1 + 2m^2 - 4m^2} = \sqrt{m^4 + 1 - 2m^2} = m^2 - 1\).
Ratios are: \( \sin \theta = \frac{m^2 - 1}{m^2 + 1} \), \( \tan \theta = \frac{m^2 - 1}{2m} \), \( \text{cosec } \theta = \frac{m^2 + 1}{m^2 - 1} \), \( \sec \theta = \frac{m^2 + 1}{2m} \), \( \cot \theta = \frac{2m}{m^2 - 1} \).

Question. Find A and B, if \(\sin (A + B) = 1\) and \(\cos (A - B) = 1\)
Answer: \( \sin (A + B) = \sin 90^\circ \implies A + B = 90^\circ \) ...(i)
\( \cos (A - B) = \cos 0^\circ \implies A - B = 0^\circ \) ...(ii)
Adding (i) and (ii): \( 2A = 90^\circ \implies A = 45^\circ \).
From (ii), \( B = A = 45^\circ \).

Question. Given that \(\sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B\), find the value of \(\sin 75^\circ\).
Answer: Let \( A = 45^\circ \) and \( B = 30^\circ \).
\( \sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side.
Answer: For an equilateral triangle inscribed in a circle of radius \( R \), the side \( a = R\sqrt{3} \).
\( a = 6\sqrt{3} \) cm.

Question. In a rectangle \(ABCD\), \(AB = 30\) cm, \(\angle BAC = 60^\circ\), calculate side \(BC\).
Answer: In \(\Delta ABC\), \(\angle B = 90^\circ\).
\( \tan 60^\circ = \frac{BC}{AB} \implies \sqrt{3} = \frac{BC}{30} \implies BC = 30\sqrt{3} \) cm.

Question. Prove the identity: \(\tan^2 A \sec^2 B - \sec^2 A \tan^2 B = \tan^2 A - \tan^2 B\)
Answer: LHS \( = \tan^2 A(1 + \tan^2 B) - (1 + \tan^2 A) \tan^2 B \)
\( = \tan^2 A + \tan^2 A \tan^2 B - \tan^2 B - \tan^2 A \tan^2 B \)
\( = \tan^2 A - \tan^2 B = \) RHS.

Question. Prove the identity: \(\sec^4 A (1 - \sin^4 A) - 2 \tan^2 A = 1\)
Answer: LHS \( = \sec^4 A (1 - \sin^2 A)(1 + \sin^2 A) - 2 \tan^2 A \)
\( = \sec^4 A \cdot \cos^2 A (1 + \sin^2 A) - 2 \tan^2 A \)
\( = \sec^2 A (1 + \sin^2 A) - 2 \tan^2 A = \sec^2 A + \sec^2 A \sin^2 A - 2 \tan^2 A \)
\( = \sec^2 A + \tan^2 A - 2 \tan^2 A = \sec^2 A - \tan^2 A = 1 = \) RHS.

Question. Prove the identity: \(\cot^2 A \text{cosec}^2 B - \cot^2 B \cdot \text{cosec}^2 A = \cot^2 A - \cot^2 B\)
Answer: LHS \( = \cot^2 A(1 + \cot^2 B) - \cot^2 B(1 + \cot^2 A) \)
\( = \cot^2 A + \cot^2 A \cot^2 B - \cot^2 B - \cot^2 B \cot^2 A \)
\( = \cot^2 A - \cot^2 B = \) RHS.

Question. Prove that : \(\frac{1}{\text{cosec } A - \cot A} + \frac{1}{\text{cosec } B - \cot B} + \frac{1}{\text{cosec } C - \cot C} = \text{cosec } A + \text{cosec } B + \text{cosec } C + \cot A + \cot B + \cot C\)
Answer: Each term is of the form \( \frac{1}{\text{cosec } \theta - \cot \theta} \).
Rationalizing: \( \frac{1}{\text{cosec } \theta - \cot \theta} \times \frac{\text{cosec } \theta + \cot \theta}{\text{cosec } \theta + \cot \theta} = \frac{\text{cosec } \theta + \cot \theta}{\text{cosec}^2 \theta - \cot^2 \theta} = \text{cosec } \theta + \cot \theta \).
Summing the terms for A, B, and C gives the RHS.

Question. If \(\text{cosec } A + \cot A = m\) and \(\text{cosec } A - \cot A = n\), prove that \(mn = 1\).
Answer: \( mn = (\text{cosec } A + \cot A)(\text{cosec } A - \cot A) = \text{cosec}^2 A - \cot^2 A = 1 \).

Question. If \( \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = \frac{5}{3} \), find the value of \( \frac{7\tan \theta + 2}{2\tan \theta + 7} \).
Answer: Dividing numerator and denominator of LHS by \(\cos \theta\):
\( \frac{\tan \theta + 1}{\tan \theta - 1} = \frac{5}{3} \implies 3\tan \theta + 3 = 5\tan \theta - 5 \implies 2\tan \theta = 8 \implies \tan \theta = 4 \).
Value \( = \frac{7(4) + 2}{2(4) + 7} = \frac{30}{15} = 2 \).

Question. Show that \( \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + 1} = \frac{1 + \cos A}{\sin A} \).
Answer: LHS \( = \frac{\cot A + \text{cosec } A - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1} \)
\( = \frac{(\text{cosec } A + \cot A) [1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1} = \text{cosec } A + \cot A \)
\( = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1 + \cos A}{\sin A} = \) RHS.

Question. If A and B are acute angles and \(\tan A = 1\), \(\sin B = \frac{1}{\sqrt{2}}\), find the value of \(\cos (A + B)\).
Answer: \( \tan A = 1 \implies A = 45^\circ \).
\( \sin B = \frac{1}{\sqrt{2}} \implies B = 45^\circ \).
\( \cos (A + B) = \cos (45^\circ + 45^\circ) = \cos 90^\circ = 0 \).

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

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

Question. Prove that : \((1 + \cot A + \tan A) (\sin A - \cos A) = \sin A \tan A - \cot A \cdot \cos A\).
Answer: Converting to sin and cos:
\( \left( 1 + \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A} \right) (\sin A - \cos A) = \left( \frac{\sin A \cos A + \cos^2 A + \sin^2 A}{\sin A \cos A} \right) (\sin A - \cos A) \)
\( = \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} = \frac{\sin^2 A}{\cos A} - \frac{\cos^2 A}{\sin A} = \sin A \tan A - \cot A \cos A = \) RHS.

Question. If \(\cos \theta - \sin \theta = \sqrt{2} \sin \theta\), prove that \(\cos \theta + \sin \theta = \sqrt{2} \cos \theta\).
Answer: \( \cos \theta = (\sqrt{2} + 1) \sin \theta \)
Multiplying by \( (\sqrt{2} - 1) \):
\( (\sqrt{2} - 1) \cos \theta = (\sqrt{2} - 1)(\sqrt{2} + 1) \sin \theta \)
\( \sqrt{2} \cos \theta - \cos \theta = (2 - 1) \sin \theta = \sin \theta \)
\( \implies \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)

Question. If \(\text{cosec } (A - B) = 2, \cot (A + B) = \frac{1}{\sqrt{3}}\), \( 0^\circ < (A + B) \le 90^\circ, A > B \), then find A and B.
Answer: \( \text{cosec } (A - B) = \text{cosec } 30^\circ \implies A - B = 30^\circ \) ...(i)
\( \cot (A + B) = \cot 60^\circ \implies A + B = 60^\circ \) ...(ii)
Adding: \( 2A = 90^\circ \implies A = 45^\circ \).
From (ii), \( B = 15^\circ \).

Question. Prove the identity: \(\frac{(1 + \cot A + \tan A)(\sin A - \cos A)}{\sec^3 A - \text{cosec}^3 A} = \sin^2 A \cdot \cos^2 A\)
Answer: LHS \( = \frac{\left( \frac{\sin A \cos A + \cos^2 A + \sin^2 A}{\sin A \cos A} \right) (\sin A - \cos A)}{\frac{1}{\cos^3 A} - \frac{1}{\sin^3 A}} \)
\( = \frac{\frac{\sin^3 A - \cos^3 A}{\sin A \cos A}}{\frac{\sin^3 A - \cos^3 A}{\sin^3 A \cos^3 A}} = \frac{\sin^3 A \cos^3 A}{\sin A \cos A} = \sin^2 A \cos^2 A = \) RHS.

Question. Prove the identity: \(\frac{\tan^3 \theta}{1 + \tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} = \sec \theta \cdot \text{cosec } \theta - 2 \sin \theta \cos \theta\)
Answer: LHS \( = \frac{\tan^3 \theta}{\sec^2 \theta} + \frac{\cot^3 \theta}{\text{cosec}^2 \theta} = \frac{\sin^3 \theta}{\cos^3 \theta} \cdot \cos^2 \theta + \frac{\cos^3 \theta}{\sin^3 \theta} \cdot \sin^2 \theta \)
\( = \frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta} = \frac{\sin^4 \theta + \cos^4 \theta}{\sin \theta \cos \theta} \)
\( = \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} \)
\( = \frac{1}{\sin \theta \cos \theta} - \frac{2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \sec \theta \text{cosec } \theta - 2\sin \theta \cos \theta = \) RHS.