Question

If A and B are acute angles such that
A + B and A - B satisfy the equation
tan- 4 tan 0+1=0,
(A) A=1/4
(B) A=76
(C) B=1/4
(D) B = 1/6

Answer 1

Answer:

As tan(A+B),tan(A−B) roots of tan

2

θ−4tanθ+1=0

Gives

tan(A+B)+tan(A−B)=4 ...(1)

tan(A+B).tan(A−B)=1 ...(2)

tan((A+B)+(A−B))=

1−tan(A+B).tan(A−B)

tan(A+B)+tan(A−B)

tan2A=

1−1

4

⇒2A=

2

π

⇒A=

4

π

Now from (1)

tan(A+B)+tan(A−B)=4

⇒

1−tanAtanB

tanA+tanB

+

1+tanAtanB

tanA−tanB

=4

⇒

1−tanB

1+tanB

+

1+tanB

1−tanB

=4

⇒

1−tan

2

B

(1+tanB)

2

+(1−tanB)

2

=4

⇒

1−tan

2

B

1+tan

2

B+2tanB+1+tan

2

B−2tanB

=4

⇒

1−tan

2

B

1+tan

2

B

=4⇒

cos2B

1

=2⇒cos2B=

2

1

⇒2B=

3

π

⇒B=

6

π

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