CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 07

Integrated (mixed) Questions

Question. If triangle PQR is a right triangle right angled at Q, then PQ is known as
(a) side adjacent to angle P
(b) hypotenuse
(c) side opposite to angle P
(d) side adjacent to angle R
Answer: (a) side adjacent to angle P

Question. \( 2 \cos^2 30^\circ - 1 \) is equal to
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sec 60°
Answer: (b) cos 60°

Question. Which of the following is incorrect (\( \theta \) is acute angle)?
(a) tan \( \theta \) = 3
(b) sin \( \theta \) = 3
(c) sec \( \theta \) = 3
(d) cot \( \theta \) = 3
Answer: (b) sin \( \theta \) = 3

Question. P and Q are acute angles such that P > Q. Which of the following is DEFINITELY true? 
(a) sin P < sin Q
(b) tan P > tan Q
(c) cos P > cos Q
(d) cos P > sin Q
Answer: (b) tan P > tan Q

Question. If \( \text{cosec } \theta - \cot \theta = \frac{1}{2}, 0 < \theta < \frac{\pi}{2} \), then \( \cos \theta = \)
(a) \( \frac{5}{3} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{4}{5} \)
Answer: (b) \( \frac{3}{5} \)

Question. If \( x = r \sin \theta \cdot \cos \phi, y = r \sin \theta \cdot \sin \phi \) and \( z = r \cos \theta \), then the value of \( x^2 + y^2 + z^2 \) is independent of
(a) r, \( \theta \)
(b) r, \( \phi \)
(c) \( \theta, \phi \)
(d) r
Answer: (c) \( \theta, \phi \)

Question. If \( \sin \theta \) and \( \cos \theta \) are the roots of the equation \( ax^2 - bx + c = 0 \), then \( a, b, c \) satisfy the relation
(a) \( b^2 - a^2 = 2ac \)
(b) \( a^2 - b^2 = 2ac \)
(c) \( a^2 + b^2 = c^2 \)
(d) \( a^2 + b^2 = 2ac \)
Answer: (a) \( b^2 - a^2 = 2ac \)

Question. If \( \cot \theta = \frac{1}{\sqrt{3}} \), the value of \( \sec^2 \theta + \text{cosec}^2 \theta \) is
(a) 1
(b) \( \frac{40}{9} \)
(c) \( \frac{38}{9} \)
(d) \( 5\frac{1}{3} \)
Answer: (d) \( 5\frac{1}{3} \)

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

Question. If \( \tan^2 \theta + \cot^2 \theta = 2, \theta \) is an acute angle, then \( \tan^3 \theta + \cot^3 \theta \) is equal to
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (b) 2

Question. If in a triangle ABC, A and B are complementary then tan C is
(a) \( \infty \)
(b) 0
(c) 1
(d) \( \sqrt{3} \)
Answer: (a) \( \infty \)

Question. \( \frac{\sin^4 \theta - \cos^4 \theta}{\sin^2 \theta - \cos^2 \theta} \) is equal to
(a) -1
(b) 2
(c) 0
(d) 1
Answer: (d) 1

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

Question. If \( \sec \alpha + \tan \alpha = m \), then \( \sec^4 \alpha - \tan^4 \alpha - 2\sec \alpha \tan \alpha \) is equal to
(a) \( m^2 \)
(b) \( -m^2 \)
(c) \( \frac{1}{m^2} \)
(d) \( \frac{-1}{m^2} \)
Answer: (c) \( \frac{1}{m^2} \)

Question. If \( \sin^4 \theta - \cos^4 \theta = k^4 \), then \( \sin^2 \theta - \cos^2 \theta \) is equal to
(a) \( k^4 \)
(b) \( k^3 \)
(c) \( k^2 \)
(d) \( k \)
Answer: (a) \( k^4 \)

Question. For all values of \( \theta, 1 + \cos \theta \) can be (\( \theta \) is acute)
(a) positive
(b) negative
(c) non-positive
(d) non-negative
Answer: (a) positive

Question. The simplified value of \( (\text{cosec } A - \sin A) (\sec A - \cos A) (\tan A + \cot A) \) is
(a) -1
(b) 2
(c) 0
(d) 1
Answer: (d) 1

Question. If \( x = a(\text{cosec } \theta + \cot \theta) \) and \( y = b(\cot \theta - \text{cosec } \theta) \) then
(a) \( xy - ab = 0 \)
(b) \( xy + ab = 0 \)
(c) \( \frac{x}{a} + \frac{y}{b} = 1 \)
(d) \( x^2y^2 = ab \)
Answer: (b) \( xy + ab = 0 \)

Question. \( \frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta} \) is equal to
(a) 2 sec² \( \theta \)
(b) 2 cos² \( \theta \)
(c) 0
(d) 1
Answer: (a) 2 sec² \( \theta \)

Question. If \( \cos y = 0 \), then what is the value of \( \frac{1}{2} \cos \frac{y}{2} \)? 
(a) 0
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{\sqrt{2}} \)
(d) \( \frac{1}{2\sqrt{2}} \)
Answer: (d) \( \frac{1}{2\sqrt{2}} \)

Question. If \( \frac{\sin^2 \alpha - 3 \sin \alpha + 2}{\cos^2 \alpha} = 1 \), then \( \alpha \) can be
(a) 60°
(b) 45°
(c) 0°
(d) 30°
Answer: (d) 30°

Question. If \( \frac{1 + \sin \alpha}{1 - \sin \alpha} = \frac{m^2}{n^2} \), then \( \sin \alpha \) is equal to
(a) \( \frac{m^2 + n^2}{m^2 - n^2} \)
(b) \( \frac{m^2 - n^2}{m^2 + n^2} \)
(c) \( \frac{m^2 + n^2}{n^2 - m^2} \)
(d) \( \frac{n^2 - m^2}{m^2 + n^2} \)
Answer: (b) \( \frac{m^2 - n^2}{m^2 + n^2} \)

Question. If \( \sin \theta - \cos \theta = \frac{3}{5} \), then \( \sin \theta \cos \theta \) is equal to
(a) \( \frac{16}{25} \)
(b) \( \frac{9}{16} \)
(c) \( \frac{9}{25} \)
(d) \( \frac{8}{25} \)
Answer: (d) \( \frac{8}{25} \)

Question. In a right-angled triangle PQR, \( \angle Q = 90^\circ \). Which of these is ALWAYS 0?
(a) cos P – sec R
(b) tan P – cot R
(c) sin P – cosec R
(d) Cannot be known without knowing the value of P
Answer: (b) tan P – cot R

Question. Evaluate : \( \frac{\tan^2 60^\circ + 4\sin^2 45^\circ + 3\sec^2 30^\circ + 5\cos^2 90^\circ}{\text{cosec } 30^\circ + \sec 60^\circ - \cot^2 30^\circ} \)
Answer: \( \text{Numerator} = (\sqrt{3})^2 + 4 \left(\frac{1}{\sqrt{2}}\right)^2 + 3 \left(\frac{2}{\sqrt{3}}\right)^2 + 5(0)^2 = 3 + 2 + 4 = 9 \).
\( \text{Denominator} = 2 + 2 - (\sqrt{3})^2 = 4 - 3 = 1 \).
The value is \( \frac{9}{1} = 9 \).

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

Question. If \( \frac{\cos \alpha}{\cos \beta} = m \) and \( \frac{\cos \alpha}{\sin \beta} = n \), show that \( (m^2 + n^2) \cos^2 \beta = n^2 \). 
Answer: LHS \( = \left( \frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta} \right) \cos^2 \beta \)
\( = \cos^2 \alpha \left( \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \right) \cos^2 \beta \)
\( = \frac{\cos^2 \alpha \cdot 1}{\sin^2 \beta} = n^2 = \text{RHS} \).

Question. In a \( \Delta ABC \) right angled at C, if \( \tan A = \frac{1}{\sqrt{3}} \) and \( \tan B = \sqrt{3} \), show that \( \sin A \cos B + \cos A \sin B = 1 \).
Answer: \( \tan A = \frac{1}{\sqrt{3}} \implies A = 30^\circ \). Since \( \angle C = 90^\circ \), \( B = 60^\circ \).
LHS \( = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ \)
\( = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 = \text{RHS} \).

Question. Find the value of x if \( \cos x = \cos 60^\circ \cdot \cos 30^\circ + \sin 60^\circ \cdot \sin 30^\circ \).
Answer: \( \cos x = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \).
\( \implies x = 30^\circ \).

Question. The altitude AD of a \( \Delta ABC \) in which \( \angle A \) is obtuse, is 10cm. If BD = 10 cm and CD = \( 10\sqrt{3} \) cm, determine \( \angle A \).
Answer: In right \( \Delta ADB \), \( \tan \angle DAB = \frac{BD}{AD} = \frac{10}{10} = 1 \implies \angle DAB = 45^\circ \).
In right \( \Delta ADC \), \( \tan \angle DAC = \frac{CD}{AD} = \frac{10\sqrt{3}}{10} = \sqrt{3} \implies \angle DAC = 60^\circ \).
\( \therefore \angle A = \angle DAB + \angle DAC = 45^\circ + 60^\circ = 105^\circ \).

Question. Using the formula \( \cos 2\theta = 2 \cos^2 \theta - 1 \), find the value of cos 30°, it is being given that \( \cos 60^\circ = \frac{1}{2} \).
Answer: Putting \( 2\theta = 60^\circ \implies \theta = 30^\circ \).
\( \cos 60^\circ = 2 \cos^2 30^\circ - 1 \)
\( \implies \frac{1}{2} = 2 \cos^2 30^\circ - 1 \implies 2 \cos^2 30^\circ = \frac{3}{2} \implies \cos^2 30^\circ = \frac{3}{4} \)
\( \implies \cos 30^\circ = \frac{\sqrt{3}}{2} \).

Question. Using the formula \( \cos 2\theta = 1 - 2 \sin^2 \theta \), find the value of sin 30°, if \( \cos 60^\circ = 1/2 \).
Answer: Putting \( 2\theta = 60^\circ \implies \theta = 30^\circ \).
\( \cos 60^\circ = 1 - 2 \sin^2 30^\circ \)
\( \implies \frac{1}{2} = 1 - 2 \sin^2 30^\circ \implies 2 \sin^2 30^\circ = \frac{1}{2} \implies \sin^2 30^\circ = \frac{1}{4} \)
\( \implies \sin 30^\circ = \frac{1}{2} \).

Question. If \( \cos \alpha = \frac{1}{2} \) and \( \tan \beta = \frac{1}{\sqrt{3}} \), find \( \sin (\alpha + \beta) \) where \( \alpha \) and \( \beta \) both are acute angles.
Answer: \( \cos \alpha = 1/2 \implies \alpha = 60^\circ \). \( \tan \beta = 1/\sqrt{3} \implies \beta = 30^\circ \).
\( \sin (\alpha + \beta) = \sin (60^\circ + 30^\circ) = \sin 90^\circ = 1 \).

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

Question. Prove that : \( \frac{\cot^2 A (\sec A - 1)}{1 + \sin A} = \sec^2 A \left( \frac{1 - \sin A}{1 + \sec A} \right) \).
Answer: LHS \( = \frac{\frac{\cos^2 A}{\sin^2 A} (\sec A - 1)}{1 + \sin A} = \frac{\cos^2 A \cdot (\sec A - 1)}{(1 - \cos^2 A)(1 + \sin A)} \). This way is complex.
Alternatively, use \( \frac{1 - \sin A}{1 - \sin A} \) for rationalization.
LHS \( = \frac{\cot^2 A (\sec A - 1)(1 - \sin A)}{1 - \sin^2 A} = \frac{\cot^2 A (\sec A - 1)(1 - \sin A)}{\cos^2 A} \)
\( = \frac{1}{\sin^2 A} (\sec A - 1)(1 - \sin A) \).
RHS \( = \sec^2 A \cdot \frac{(1 - \sin A)(\sec A - 1)}{\sec^2 A - 1} = \frac{\sec^2 A (1 - \sin A)(\sec A - 1)}{\tan^2 A} \)
\( = \frac{1}{\cos^2 A} \cdot \frac{\cos^2 A}{\sin^2 A} (1 - \sin A)(\sec A - 1) = \frac{1}{\sin^2 A} (1 - \sin A)(\sec A - 1) = \text{LHS} \).

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

Question. If \( \cot \theta + \tan \theta = x \) and \( \sec \theta - \cos \theta = y \), prove that \( (x^2y)^{2/3} - (xy^2)^{2/3} = 1 \). 
Answer: \( x = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1}{\sin \theta \cos \theta} \). \( y = \frac{1}{\cos \theta} - \cos \theta = \frac{\sin^2 \theta}{\cos \theta} \).
\( x^2 y = \frac{1}{\sin^2 \theta \cos^2 \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} = \frac{1}{\cos^3 \theta} \implies (x^2 y)^{2/3} = \frac{1}{\cos^2 \theta} = \sec^2 \theta \).
\( x y^2 = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin^4 \theta}{\cos^2 \theta} = \frac{\sin^3 \theta}{\cos^3 \theta} = \tan^3 \theta \implies (x y^2)^{2/3} = \tan^2 \theta \).
\( \therefore \sec^2 \theta - \tan^2 \theta = 1 \). Proved.

Question. If \( \text{cosec } \theta - \sin \theta = a^3 \), \( \sec \theta - \cos \theta = b^3 \), prove that \( a^2 b^2 (a^2 + b^2) = 1 \). 
Answer: \( a^3 = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \implies a = \frac{\cos^{2/3} \theta}{\sin^{1/3} \theta} \).
\( b^3 = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \implies b = \frac{\sin^{2/3} \theta}{\cos^{1/3} \theta} \).
\( a^2 b^2 = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} \cdot \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} = \cos^{2/3} \theta \sin^{2/3} \theta \).
\( a^2 + b^2 = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} + \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin^{2/3} \theta \cos^{2/3} \theta} = \frac{1}{\sin^{2/3} \theta \cos^{2/3} \theta} \).
LHS \( = (\cos^{2/3} \theta \sin^{2/3} \theta) \cdot \frac{1}{\sin^{2/3} \theta \cos^{2/3} \theta} = 1 = \text{RHS} \).

Question. If \( a \cos^3 \theta + 3a \cos \theta \sin^2 \theta = m, a \sin^3 \theta + 3a \cos^2 \theta \sin \theta = n \), prove that \( (m + n)^{2/3} + (m - n)^{2/3} = 2a^{2/3} \). 
Answer: \( m + n = a(\cos^3 \theta + 3 \cos \theta \sin^2 \theta + 3 \cos^2 \theta \sin \theta + \sin^3 \theta) = a(\cos \theta + \sin \theta)^3 \).
\( m - n = a(\cos \theta - \sin \theta)^3 \).
LHS \( = [a(\cos \theta + \sin \theta)^3]^{2/3} + [a(\cos \theta - \sin \theta)^3]^{2/3} \)
\( = a^{2/3} (\cos \theta + \sin \theta)^2 + a^{2/3} (\cos \theta - \sin \theta)^2 \)
\( = a^{2/3} (1 + 2 \sin \theta \cos \theta + 1 - 2 \sin \theta \cos \theta) = 2a^{2/3} = \text{RHS} \).

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

Question. Prove that : \( \frac{1}{(\cot A)(\sec A) - \cot A} - \text{cosec } A = \text{cosec } A - \frac{1}{(\cot A)(\sec A) + \cot A} \)
Answer: Note: \( (\cot A)(\sec A) = \frac{\cos A}{\sin A} \cdot \frac{1}{\cos A} = \frac{1}{\sin A} = \text{cosec } A \).
Question becomes: \( \frac{1}{\text{cosec } A - \cot A} - \text{cosec } A = \text{cosec } A - \frac{1}{\text{cosec } A + \cot A} \)
LHS \( = \frac{\text{cosec } A + \cot A}{\text{cosec}^2 A - \cot^2 A} - \text{cosec } A = (\text{cosec } A + \cot A) - \text{cosec } A = \cot A \).
RHS \( = \text{cosec } A - \frac{\text{cosec } A - \cot A}{\text{cosec}^2 A - \cot^2 A} = \text{cosec } A - (\text{cosec } A - \cot A) = \cot A \).
LHS = RHS. Proved.

Question. Prove that: \( \sin^6 A + 3 \sin^2 A \cos^2 A = 1 - \cos^6 A \)
Answer: LHS \( = (\sin^2 A)^3 + 3 \sin^2 A \cos^2 A \cdot 1 \)
\( = (1 - \cos^2 A)^3 + 3(1 - \cos^2 A) \cos^2 A \)
\( = 1 - \cos^6 A - 3 \cos^2 A (1 - \cos^2 A) + 3 \cos^2 A - 3 \cos^4 A \)
\( = 1 - \cos^6 A - 3 \cos^2 A + 3 \cos^4 A + 3 \cos^2 A - 3 \cos^4 A = 1 - \cos^6 A = \text{RHS} \).

Question. If \( \frac{1}{\sin \theta - \cos \theta} = \frac{\text{cosec } \theta}{\sqrt{2}} \), prove that \( \left( \frac{1}{\sin \theta + \cos \theta} \right)^2 = \frac{\sec^2 \theta}{2} \). 
Answer: \( \frac{1}{\sin \theta - \cos \theta} = \frac{1}{\sqrt{2} \sin \theta} \implies \sqrt{2} \sin \theta = \sin \theta - \cos \theta \implies \cos \theta = (1 - \sqrt{2}) \sin \theta \).
This implies \( \cot \theta = 1 - \sqrt{2} \).
Alternatively: \( \sin \theta - \cos \theta = \sqrt{2} \sin \theta \implies -\cos \theta = (\sqrt{2} - 1) \sin \theta \implies -1 = (\sqrt{2} - 1) \tan \theta \).
Then \( \tan \theta = \frac{-1}{\sqrt{2} - 1} = -(\sqrt{2} + 1) \).
\( (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + 2 \frac{\tan \theta}{1 + \tan^2 \theta} \). This is complex.
Better: \( \sin \theta - \cos \theta = \sqrt{2} \sin \theta \implies \cos \theta = (1 - \sqrt{2}) \sin \theta \).
Square: \( \cos^2 \theta = (1 + 2 - 2\sqrt{2}) \sin^2 \theta = (3 - 2\sqrt{2}) \sin^2 \theta \).
LHS \( = \frac{1}{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta} = \frac{1/\cos^2 \theta}{1 + \tan^2 \theta + 2 \tan \theta} = \frac{\sec^2 \theta}{(1 + \tan \theta)^2} \).
From \( 1 = (1 - \sqrt{2}) \tan \theta \implies \tan \theta = \frac{1}{1 - \sqrt{2}} = -(1 + \sqrt{2}) \).
Then \( 1 + \tan \theta = 1 - 1 - \sqrt{2} = -\sqrt{2} \).
\( \therefore \frac{\sec^2 \theta}{(-\sqrt{2})^2} = \frac{\sec^2 \theta}{2} = \text{RHS} \).

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): The value of each of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Reason (R): In right \( \Delta ABC \), \( \angle B = 90^\circ \) and \( \angle A = \theta \), \( \sin \theta = \frac{BC}{AC} < 1 \) and \( \cos \theta = \frac{AB}{AC} < 1 \) as hypotenuse is the longest side.
Answer: (b) Both A and R are true but R is not the correct explanation of A.

Question. Assertion (A): In a \( \Delta ABC \), right angled at B, if \( \sin A = \frac{8}{17} \), then \( \cos A = \frac{15}{17} \) and \( \tan A = \frac{8}{15} \).
Reason (R): For acute angle \( \theta \), \( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \) and \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \).
Answer: (a) Both A and R are true and R is the correct explanation of A.