if cosA=-12÷13andcot B=24÷7then prove that sin(A+B)=-36\325 pls give me the ans
Answers
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 Read 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?show=654590#a654590