Question

If sin A=1/3, then find the value of 9cot square A + 9.​

Answer 1

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

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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