If $117 cos^2 A + 129 sin^2A = 120 $ and $170 cos^2 B + 158sin^2 B = 161 $, then the value of $cosec^2A\, sec^2B$ is :
Answer & explanation
Correct answer: option 4
117cos²A + 129sin²A = 120
117cos²A + 117sin²A + 12sin²A = 120
{ cos²A + sin²A = 1 }
12sin²A = 120 - 117 = 3
sinA = \(\frac{1 }{2}\)
cosecA = 2
Now,
170 cos²B + 158 sin²B = 161
158 cos²B + 12 cos²B + 158 sin²B = 161
12 cos²B = 3
cosB = \(\frac{1 }{2}\)
secB = 2
Now, cosec²A. sec²B
= (2)² . (2)²
= 16