If sinA, sinB, cosA are in G.P., then roots of x2 + 2xcotB+1=o are always
a equal
b real
c imaginary
d greatee than 1
Answers
Answered by
15
Answer:
sina,sinb,cosa are in G.P.
∴ sin2b=sinacosa -------- ( 1 )
⇒ x2+2xcotb+1=0
⇒ D=(2cotb)2−4(1)(1)
=4cot2b−1
=4(sin2bcos2b−1)
=4(sin2bcos2b−sin2b)
=4(sin2b1−sin2b−sin2b)
=4(sin2b1−2sin2b)
=4(sin2b1−2sinacosa) [ From equation ( 1) ]
=4(sin2bsin2a+cos2a−2sinacosa)
=4(sin2b(sina−cosb)2)
⇒ D=(sinb2(sina−cosa))2≥0
∴ Roots of given equation are real.
Step-by-step explanation:
Answered by
4
Answer:
your answer is b) real
Step-by-step explanation:
hope it's help you
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