Question

Find tanA if cosA=-12/13 and A lies in the third quadrant.

Answer 1

Answer:

Given as cos A = -12/13 and cot B = 24/7 As we know that, A lies in second quadrant, B in the third quadrant. Since, in the second quadrant sine function is positive. Since, in the third quadrant, both sine and cosine functions are negative. On using the formulas, sin A = √(1 – cos2 A), sin B = – 1/√(1 + cot2 B) and cos B = -√(1 – sin2 B), Therefore let us find the value of sin A and sin B sin A = √(1 – cos2 A) = √(1 – (-12/13)2) = √(1 – 144/169) = √((169 - 144)/169) = √(25/169) = 5/13 sin B = – 1/√(1 + cot2 B) = – 1/√(1 + (24/7)2) = – 1/√(1 + 576/49) = -1/√((49 + 576)/49) = -1/√(625/49) = -1/(25/7) = -7/25 cos B = -√(1 – sin2 B) = -√(1 - (-7/25)2) = -√(1 - (49/625)) = -√((625 - 49)/625) = -√(576/625) = -24/25 Therefore, now let us find (i) sin (A + B) As we know that sin (A + B) = sin A cos B + cos A sin B Therefore, sin (A + B) = sin A cos B + cos A sin B = 5/13 × (-24/25) + (-12/13) × (-7/25) = -120/325 + 84/325 = -36/325 (ii) cos (A + B) As we know that cos (A + B) = cos A cos B – sin A sin B Therefore, cos (A + B) = cos A cos B – sin A sin B = -12/13 × (-24/25) – (5/13) × (-7/25) = 288/325 + 35/325 = 323/325 (iii) tan (A + B) As we know that tan (A + B) = sin (A + B)/cos (A + B) = (-36/325)/(323/325) = -36/323Read more on Sarthaks.com - https://www.sarthaks.com/654582/cos-cot-where-lies-the-second-quadrant-and-the-third-quadrant-find-the-values-the-following

Answer 2

Answer:

Answer is 5/12

hope it works

like my answer