© examsiri.com
Question : 14 of 150
Marks:
+1,
-0
Solution:
Since
sin‌A,sin‌B and
cos‌A are in GP
∴sin2B=sin‌A‌cos‌A ......(i)
x2+2x‌cot‌B+1=0 (given)
Now,
b2−4ac=4cot2B−4 = =| 4(1−sin2B)−4sin2B |
| sin2B |
= =| 4[1−2‌sin‌A‌cos‌A] |
| sin2B |
[from (i)]
=4()2>0 ∴ Roots are always real.
© examsiri.com
Go to Question: