Question

In triangle ABC, tan( A-B-C)=​

Answer 1

Answer:

A+B+C=180 so that

B+C=180-A

A-B-C=A-(B+C)=A-(180-A)=2A-180

tan(x-y)=(tanx-tany)/(1+tanxtany)

Therefore ,tan(2A-180)=(tan2A-tan(180))/(1+tan2Atan(180))

=tan2A/1-tan2A)

since tan(180)=-1

Answer 2

Concept:

We need the concept of trigonometry.

Trigonometry is the branch of mathematics that deals with the sides and angles of triangles using functions like sine, cosine, tangent and cot etc.

Given:

We are given triangle ABC and tan(A-B-C) where A, B and C are angles.

To find:

We are asked to find the value of tan(A-B-C)

Solution:

In a triangle ABC, sum of its angles is equal to 180°

So, A+B+C=180°

⇒B+C=180°-A ....................................(1)

Therefore,

tan(A-B-C)

=tan(A-(B+C))

=tan(A-(180°-A)) (using equation 1)

=tan(A+A-180°)

=tan(2A-180°)

=(tan2A - tan 180°)/(1+tan2Atan180°)  (expanding using the formula for                 tan(A-B)=(tan A-tan B)/1+tanAtanB )

=(tan2A-0)/(1+tan2Ax0)  (since tan 180°=0)

=tan2A

Hence,

tan(A-B-C)=tan2A