tana, tanb are roots of the equation x^2-6x+c=0 the value of sin^2(a+b) is
Answers
Answer:
Correct option is
A
tanA=
b
a
B
tanB=
a
b
D
tanA+tanA=
ab
c
2
If tanA,tanB are the roots of the quadratic abx
2
−c
2
x+ab=0 then
tanA+tanB=
ab
c
2
tanAtanB=1
∵tan(A+B)=
1−tanAtanB
tanA+tanB
=
1−1
ab
c
2
=
0
c
2
=∞
∴A+B=
2
π
∴C=
2
π
∴cosC=
2ab
a
2
+b
2
−c
2
=∞
∴a
2
+b
2
=c
2
Using sine rule, we have
⇒(2RsinA)
2
+(2RsinB)
2
=(2RsinC)
2
⇒4R
2
sin
2
A+4R
2
sin
2
B=4R
2
sin
2
C
Dividing both sides by 4R
2
we get
⇒sin
2
A+sin
2
B=sin
2
C
Add sin
2
C to both sides, we get
⇒sin
2
A+sin
2
B+sin
2
C=2sin
2
C
Since C=
2
π
(from above)
⇒sin
2
A+sin
2
B+sin
2
C=2sin
2
2
π
⇒sin
2
A+sin
2
B+sin
2
C=2
∴tanA+tanB=
ab
c
2
=
ab
a
2
+b
2
=
b
a
+
a
b
But, tanAtanB=1=
b
a
a
b
∴tanA=
b
a
,tanB=
a
b
Step-by-step explanation:
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