Question

in a triangle ABC right angled at A then tan B × tanC = ?​

Answer 1

Helo mate here your ans

Given: tanA:tanB:tanC=1:2:3

⇒

1

tanA

=

2

tanB

=

3

tanC

=k

⇒∑tanA=∏tanA

⇒6k=6k

3

⇒k(1−k

2

)=0

⇒k



=0 and k

2

=1

⇒k=1

∴tanA=1,tanB=2,tanC=3

⇒

cotA

1

=1,

cotB

1

=2,

cotC

1

=3

or cotA=1,cotB=

2

1

,cotC=

3

1

⇒cot

2

A=1,cot

2

B=

4

1

,cot

2

C=

9

1

Add 1 to both sides,we get

⇒1+cot

2

A=2,1+cot

2

B=1+

4

1

=

4

5

,1+cot

2

C=1+

9

1

=

9

10

We know that csc

2

θ=1+cot

2

θ

⇒csc

2

A=2

csc

2

B=

4

5

csc

2

C=

9

10

⇒sin

2

A=

2

1

,sin

2

B=

5

4

,sin

2

C=

10

9

Using sine rule, we have

⇒a

2

:b

2

:c

2

=

2

1

:

5

4

:

10

9

=5:8:9 (on simplification)

Hope this ans help you..