if sin A=1/3, then find the value of (9 cot² A+9)
Answers
Answered by
18
TRIGONOMETRIC RESOLUTION
Attachments:
Answered by
42
sinA = 1/3 = P/h
b = √( h² - P²) = √(3² -1²) = 2√2
cotA = b/P = 2√2/1 = 2√2
now,
( 9cot²A + 9)
= {9×(2√2)² + 9}
= 72 + 9
= 81
_____________________________________________________
method2 :-
sinA = 1/3
sin²A = 1/9
1/cosec²A = 1/9
cosec²A = 9
1 + cot²A = 9 [ cosec²A - cot²A = 1 ]
cot²A = 8
9cot²A= 72
9 cot²A + 9 = 81
9(cot²A + 1) = 81
b = √( h² - P²) = √(3² -1²) = 2√2
cotA = b/P = 2√2/1 = 2√2
now,
( 9cot²A + 9)
= {9×(2√2)² + 9}
= 72 + 9
= 81
_____________________________________________________
method2 :-
sinA = 1/3
sin²A = 1/9
1/cosec²A = 1/9
cosec²A = 9
1 + cot²A = 9 [ cosec²A - cot²A = 1 ]
cot²A = 8
9cot²A= 72
9 cot²A + 9 = 81
9(cot²A + 1) = 81
Similar questions