Question

Write down the
value
of a sin 3/5
119/169 = cos²x.​

Answer 1

Step-by-step explanation:

cosx = -5/13

sin²x + cos²x = 1

sin²x + (5/13)² = 1

sin²x + 25/169 = 1

sin²x = 1 - 25/169 = (169-25)/169 = 144/169

sinx = +12/13

secx = -13/5

sec²x = 1+tan²x

169/25= 1+ tan²x

(169/25) - 1 = tan²x

= (169-25)/25 = 144/25

tanx = -12/5

cosx is given as 5/13, but it is -5/13 (because it is in II quadrant, this question is not wrong.)

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