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8. त्रिकोणमिति का परिचय Mathematics class 10 in Hindi Medium ncert book solutions प्रश्नावली 8.3
8. त्रिकोणमिति का परिचय प्रश्नावली 8.3 – Complete NCERT Book Solutions for Class 10 Mathematics (Hindi Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 8. त्रिकोणमिति का परिचय प्रश्नावली 8.3 to help you master concepts and score higher.
8. त्रिकोणमिति का परिचय Mathematics class 10 in Hindi Medium ncert book solutions प्रश्नावली 8.3
NCERT Solutions for Class 10 Mathematics play an important role in helping students understand the concepts of the chapter 8. त्रिकोणमिति का परिचय clearly. This chapter includes the topic प्रश्नावली 8.3, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 10 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 8. त्रिकोणमिति का परिचय is explained in a structured way, making it easier for students to connect the theory with the topic प्रश्नावली 8.3. By studying these updated NCERT Solutions for Class 10 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
8. त्रिकोणमिति का परिचय
प्रश्नावली 8.3
अध्याय 8. त्रिकोणमितिय अनुपातों का परिचय
प्रश्नावली 8.3
Q1. निम्नलिखित का मान निकालिए:
(iii) cos 48° - sin 42°
हल: cos 48° - sin 42°
⇒ sin(90° - 48°) - sin 42°
⇒ sin 42° - sin 42° = 0
(iv) cosec 31° - sec 59°
हल: cosec 31° - sec 59°
⇒ sec (90° - 31°) - sec 59° [ cosec q = sec (90° - q) ]
⇒ sec 59° - sec 59° = 0
Q2. दिखाइए कि
(i) tan 48° tan 23° tan 42° tan 67° = 1
हल: (i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= cot (90° - 48°) tan (90° - 23°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° × tan 42°) (cot 67° × tan 67°)
= 1 × 1 [ cot A × tan A = 1 ]
= 1
LHS = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
हल: (ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° sin 38° sin 52°
= sin (90° - 38°) cos 52° – cos (90° - 38°) sin 52°
= sin 52° cos 52° - cos 52° sin 52°
= sin 52° (cos 52° - cos 52°)
= sin 52° × 0
= 0
LHS = RHS
Q3. यदि tan 2A = cot(A - 18°), जहाँ 2A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: tan 2A = cot(A - 18°),
⇒ cot (90° - 2A) = cot(A - 18°)
दोनों पक्षों में तुलना करने पर
⇒ 90° - 2A = A - 18°
⇒ 90° + 18° = A + 2A
⇒ 3A = 108°
Q4. यदि tan A = cot B, तो सिद्ध कीजिए कि A + B = 90°
हल: tan A = cot B दिया है |
⇒ tan A = tan (90° - B) तुलना करने पर
⇒ A = 90° - B
⇒ A + B = 90° Proved
Q5. यदि sec 4A = cosec(A - 20°), जहाँ 4A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: sec 4A = cosec(A - 20°)
⇒ cosec (90° - 4A) = cosec(A - 20°) [ sec q = (90°- q) ]
तुलना करने पर
⇒ 90° - 4A = A - 20°
⇒ 90° + 20° = A + 4A
⇒ 5A = 110°
Q7. sin 67° + cos 75° को 0° और 45° के बीच के कोणों के त्रिकोणमितिय अनुपातों के पदों में व्यक्त कीजिए |
हल : sin 67° + cos 75°
⇒ cos (90° - 67°) + sin (90° - 75°)
⇒ cos 23° + sin 15°
इस पाठ के अन्य दुसरे विषय भी देखे :
1. प्रश्नावली 8.1 2. प्रश्नावली 8.2 3. प्रश्नावली 8.3 4. प्रश्नावली 8.4 5. प्रश्नावली 8.5