Question

If A, B, C are interiorvangles of ABC, then show that tan B+C/2=cos C/2​

Answer 1

Answer:

Sin[(B+C)/2]

Since A+B+C=180 for interior angles of triangle ABC.

then B+C=180-A.

NOW   Sin [(180-A)/2]

             =Sin[90-(A/2)]                      since Sin(90-A)=CosA

              =Cos(A/2)

PLZ MARK AS BRIANLIEST,FLW ME AND THX FOR THE SUPERB QUESTION

Answer 2

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180

o

or A+B+C=180

o

B+C=180

o

−A

Multiply both sides by

2

1

2

1

(B+C)=

2

1

(180

o

−A)

2

1

(B+C)=90

o

−

2

A

...(1)

Now

2

1

(B+C)

Taking sine of this angle

sin(

2

B+C

)[

2

B+C

=90

o

−

2

A

]

sin(90

o

−

2

A

)

cos

2

A

[sin(90

o

−θ)=cosθ]

Hence sin(

2

B+C

)=cos

2

Aproved