Chapter 23. Trigonometric Ratios of Some Special Angles NIOS Class 10 Mathematics Chapter-Wise Important Questions and Answers

Chapter 23. Trigonometric Ratios of Some Special Angles | NIOS Class 10 Mathematics Important Questions, Fully Solved Answers, and Chapter-wise Exam Preparation Guide.

Chapter 23. TRIGONOMETRIC RATIOS OF SOME SPECIAL ANGLES

Optional Questions and Answers for Algebraic Expressions and Polynomials [1 Mark]

CHECK YOUR PROGRESS 3.1

Short Questions and Answers for Algebraic Expressions and Polynomials [1 or 2 Marks]

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Long Questions and Answers for Algebraic Expressions and Polynomials [ 3 or 4 Mark]

CHECK YOUR PROGRESS 23.1

 1. Evaluate each of the following:

(i)  sin²60° + cos²45°

(ii) 2 sin²30° – 2cos² 45° + tan²60°

(iii) 4 sin² 60° + 3 tan²30° – 8 sin² 45° cos 45°

(iv) 4sin430°+cos460°-3cos245°-2sin245°

(v)

tan45°cosec30°+sec60°cot45°-5sin90°2cos0°

(vi)

5cos²60°+4sec²30°-tan²45°sin²30°+cos²30°

Solution: (i) We have,    sin260° + cos245°

=322+122=34+12=3+24=54

 

(ii) We have,   2sin230°-2cos245°+tan260°

=2×122-2×122+32

=2×14-2×12+3 

=12-1+3=12+2=1+42=52

 

(iii) 4 sin² 60° + 3 tan²30° – 8 sin² 45° cos 45°

Solution: We have, 4 sin260° + 3 tan230°- 8sin45°cos45°

=4×322+3×132-8×12×12

=4×34+3×13-8×12 

=3+1-4=4-4=0 

 

(iv) We have,  4sin430°+cos460°-3cos245°-2sin245°

=4124+124-3122-2122 

=4116+116-312-2×12

=4216-312-1

=816-31-22

=12+32=42=2

 

 

(v)

tan45°cosec30°+sec60°cot45°-5sin90°2cos0°=

=12+21-52×11=1+42-52

=52-52=0

(vi)

5cos260°+4sec230°-tan245°sin230°+cos230°

=5×122+4×232-12122+322

=5×14+4×43-114+34

=54+163-144=15+64-12121=67 12 

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Posted 1 month ago