If sinA=-5/13 and A lies in the 3rd quadrant, then prove that 5 cot^2A+12 tanA+ 13 cosecA=0
Answers
In the 3rd quadrant, sine, cosine, secant, cosecant are -ve. Given, sinA = (-5/13).
cosA = √(1 - sin²A) = √1 - (-5/13)² = 12/13
cosA = -12/13 [A lies in 3rd quadrant]
Therefore,
(i): cotA = cosA/sinA = (-12/13)/(-5/13)
cotA = 12/5
(ii): tanA = 1/cotA = 1/(12/5)
tanA = 5/12
(iii): cosecA = 1/sinA = 1/(-5/13)
cosecA = 13/5
Hence, 5 cot²A + 12tanA + 13cosecA
= 5 (12/5)² + 12(5/12) + 13(-13/5)
= 144/5 + 5 - 169/5
= (144 + 25 - 169)/5
= 0 proved
Answer:
Step-by-step explanation:
In the 3rd quadrant, sine, cosine, secant, cosecant are -ve. Given, sinA = (-5/13).
cosA = √(1 - sin²A) = √1 - (-5/13)² = 12/13
cosA = -12/13 [A lies in 3rd quadrant]
Therefore,
(i): cotA = cosA/sinA = (-12/13)/(-5/13)
cotA = 12/5
(ii): tanA = 1/cotA = 1/(12/5)
tanA = 5/12
(iii): cosecA = 1/sinA = 1/(-5/13)
cosecA = 13/5
Hence, 5 cot²A + 12tanA + 13cosecA
= 5 (12/5)² + 12(5/12) + 13(-13/5)
= 144/5 + 5 - 169/5
= (144 + 25 - 169)/5
= 0 proved