Question

in a triangle abc if (b-c)/(b+c)cot (A/2)+(b+c)/(b-c)tan(a/2)=2 then triangle abc is which type of triangle

Answer 1

Answer:

data insufficient.

Step-by-step explanation:

Given that,    $\frac{b-c}{b+c}cot(\frac{A}{2}) + \frac{b+c}{b-c}tan(\frac{A}{2} ) = 2$

                => $tan(\frac{B-C}{2} )+ \frac{1}{tan(\frac{B-C}{2}) } = 2$

                => $tan^{2}(\frac{B-C}{2})+1=2tan(\frac{B-C}{2} )$

                => $tan^{2} (\frac{B-C}{2})-2 tan(\frac{B-C}{2}) +1=0$

                => $(tan(\frac{B-C}{2}) -1)^{2}=0$

                => $tan(\frac{B-C}{2}) = 1$

                => $B-C = \frac{\pi}{2}$

Therefore, it is not possible to say about the type of triangle.

               

Answer 2

Answer:

Given that, \frac{b-c}{b+c}cot(\frac{A}{2}) + \frac{b+c}{b-c}tan(\frac{A}{2} ) = 2

b+c

b−c

cot(

2

A

)+

b−c

b+c

tan(

2

A

)=2

=> tan(\frac{B-C}{2} )+ \frac{1}{tan(\frac{B-C}{2}) } = 2tan(

2

B−C

)+

tan(

2

B−C

)

1

=2

=> tan^{2}(\frac{B-C}{2})+1=2tan(\frac{B-C}{2} )tan

2

(

2

B−C

)+1=2tan(

2

B−C

)

=> tan^{2} (\frac{B-C}{2})-2 tan(\frac{B-C}{2}) +1=0tan

2

(

2

B−C

)−2tan(

2

B−C

)+1=0

=> (tan(\frac{B-C}{2}) -1)^{2}=0(tan(

2

B−C

)−1)

2

=0

=> tan(\frac{B-C}{2}) = 1tan(

2

B−C

)=1

=> B-C = \frac{\pi}{2}B−C=

2

π

Therefore, it is not possible to say about the type of triangle.