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Last updated at Feb. 13, 2020 by Teachoo
Transcript
Misc 4 Prove sin–1 8/17 + sin–1 3/5 = tan–1 77/36 Let a = sin–1 8/17 & b = sin–1 3/5 We convert sin–1 to tan–1 & then use tan (a – b) formula Let a = sin–1 𝟖/𝟏𝟕 sin a = 8/17 Now, cos a = √(1 −𝑠𝑖𝑛2 𝑎) = √(1 −(8/17)^2 ) = √(225/289) = 15/17 Let b = sin–1 𝟑/𝟓 sin b = 3/5 Now, cos2 b = 1 – sin2 b cos b = √(1 −𝑠𝑖𝑛2 𝑏) = √(1 −(3/5)^2 )= √(16/25) = 4/5 Let tan a = (sin 𝑎)/(cos a) = (8/17)/(15/17) = 8/17 × 17/15 = 8/15 Let tan b = sin𝑏/cos𝑏 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 Now tan (a + b) = tan〖𝑎 +〖 tan〗〖𝑏 〗 〗/(1 − tan〖𝑎 tan𝑏 〗 ) = (8/15 + 3/4)/(1 − 8/15 × 3/4) = ((8(4) + 3(15) )/(15 × 4) )/( (15 × 4 − 8 × 3)/(15 × 4) ) = ((32 + 45 )/(15 × 4) )/( (60 − 24)/(15 × 4) ) = (32 + 45)/(15 × 4) × (15 × 4)/(60 −24) = 77/36 Hence, tan (a + b) = 77/36 a + b = tan–1 77/36 Putting value of a & b sin–1 8/17 + sin–1 3/5 = tan–1 77/36 Hence Proved
Formulae based
Example 3 Important
Ex 2.2,1 Not in Syllabus - CBSE Exams 2021
Ex 2.2, 2 Important Not in Syllabus - CBSE Exams 2021
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Ex 2.2, 14 Important Not in Syllabus - CBSE Exams 2021
Example 4
Ex 2.2, 3 Important Not in Syllabus - CBSE Exams 2021
Misc. 8 Important Not in Syllabus - CBSE Exams 2021
Ex 2.2, 4 Important Not in Syllabus - CBSE Exams 2021
Ex 2.2, 15 Important Not in Syllabus - CBSE Exams 2021
Example 13 Important Not in Syllabus - CBSE Exams 2021
Misc. 14 Not in Syllabus - CBSE Exams 2021
Misc. 17 Important Not in Syllabus - CBSE Exams 2021
Misc. 13 Important Not in Syllabus - CBSE Exams 2021
Example 10 Important Not in Syllabus - CBSE Exams 2021
Misc. 3 Not in Syllabus - CBSE Exams 2021
Misc. 4 Important Not in Syllabus - CBSE Exams 2021 You are here
Misc 12 Important Not in Syllabus - CBSE Exams 2021
Misc 16 Important Not in Syllabus - CBSE Exams 2021
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