Tili Kannada Text Book Class 10 Solutions Padya Chapter 1 Eni

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KSEEB SSLC Class 10 Sanskrit नंदिनी Solutions Karnataka State Syllabus

KSEEB SSLC Class 10 Sanskrit नंदिनी Solutions Karnataka State Syllabus

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1st PUC Sanskrit Textbook Answers, Notes, Guide, Summary Pdf Download Karnataka

1st PUC Sanskrit Textbook Answers, Notes, Guide, Summary Pdf Download Karnataka

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KSEEB Solutions for Class 9 Sanskrit नंदिनी Karnataka State Syllabus

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Tili Kannada Text Book Class 7 Solutions Padya Chapter 1 Nityotsava

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Tili Kannada Text Book Class 10 Solutions Gadya Chapter 2 Asi Masi Krishi

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

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
cosec2 A – cot2 A = 1
cosec2 A = 1 + cot2 A
cosec2 A = cot2 A + 1
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 1

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
i) sin2 A + cos2 A = 1
sin2 A = 1 – cos2 A
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 2
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 3

Question 3.
Evaluate :
i) \(\frac{\sin ^{2} 63^{\circ}+\sin 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 4

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) 9 sec2 A – 9 tan2 A.
A) 1
B) 9
C) 8
D) 0
Solution:
B) 9
9 sec2 A – 9 tan2 A
= 9(sec2 A – tan2 A)
= 9 × 1
= 9

ii) (1+tan θ + sec θ) (1+ cot θ- cosec θ) =
A) 0
B) 1
C) 2
D) -1
Solution:
C) 2
(1+tan θ + sec θ) (1+ cot θ- cosec θ)
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 5
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 6

iii) (sec A + tan A) (1 – sin A) =
A) sec A
B) sin A
C) cosec A
D) cos A
Solution:
D) cos A
(sec A + tan A) (1 – sin A)
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 7

iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
A) sec2 A
B) -1
C) cot2 A
D) tan2 A
Solution:
D) tan2 A
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 8

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 9
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 10

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 11
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 12

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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 14
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Solution:
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Solution:
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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 22
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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 24
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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 27
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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 29
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Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 31
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 32

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