KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.4

Students can Download Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.4

Question 1.
Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A.
Answer:

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer:

Question 3.
Evaluate:
i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{0}+\cos ^{2} 73^{\circ}}\)
Answer:

= \(\frac{1}{1}\) = 1

ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° + cos 25°
= sin² 25° + cos² 25°
= 1. [∵ cos² θ + sin² θ = 1]

Question 4.
Choose the correct option. Justify your choice.
i) 9sec²A – 9tan²A =
A) 1
B) 9
C) 8
D) 0
= 9sec²A – 9tan²A
= 9(sec²A – tan²A)
= 9(1) = 9
therefore, correct answer is B) 9

ii) (1 + tan θ + sec θ) (1 + cotθ – cosecθ) =
A) 0
B) 1
C) 2
D) – 1
Answer:

Hence the correct answer is C) 2

iii) (sec A + tan A)(1 – sin A) =
A) sec A
B) sin A
C) cosec A
D) cos A
Answer:

= cos A
Hence the correct answer is D) cos A

iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
A) sec²A
B) – 1
C) cot²A
D) tan²A
Answer:

= tan²A
∴ correct answer is D) tan² A.

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
i) (cosecθ – cotθ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:
(cosecθ – cotθ)²

= \(\frac{1-\cos \theta}{1+\cos \theta}\)

ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2sec A
Answer:

iii) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + secθcosecθ [Hint: write the expression in terms of sin θ and cos θ]
Answer:

= cosecθ.secθ + 1

iv) \(\frac{1+\sec A}{\sec A}\) = \(\frac{\sin ^{2} A}{1-\cos A}\) [Hint: Simplify LHS and RHS separately].
Answer:
LHS

= 1 + cos A
RHS

= (1 + cos A)
∴ LHS = RHS

v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) = cosec A + cot A, using the identity cosec² A = 1 + cot² A.
Answer:


= cot A + cosec A.

vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:

vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ
Answer:

= tan θ

viii) (sinA + cosecA)² + (cos A + sec A)² = 7 + tan²A + cos²A
Answer:
(sinA + cosecA)² + (cos A + sec A)²
= sin²A + cosec²A + 2sinA cosecA + cos²A + sec²A + 2cosA secA
= sin²A + cos²A +cosec²A + sec²A + 2sinA cosecA + 2cosA secA
= sin²A + cos²A + cosec²A + sec²A + 2
sinA × \(\frac{1}{\sin A}\) + 2cos A × \(\frac{1}{\cos A}\)
= 1 + (1 + cot²A) + (1 + tan² A) + 2 + 2
= 1 + 1 + cot²A + 1 + tan²A + 2 + 2
= 7 + tan² A + cot² A.

ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\) [Hint: Simplify LHS and RHS separately].
Answer:
LHS = (cosec A – sin A)( sec A – cos A)

= cos A.sin A

∴ LHS = RHS

x)

Answer:
LHS

= (- tan A)² = tan²A
LHS = RHS