Question

IF AB, AC are the equal sides of an isosceles triangle, prove that tan A/2=cotB.​

Answer 1

Answer:

$tan\frac{A}{2}=cotB$

Step-by-step explanation:

Given:

∆ABC is an isoceles triangle with AB = AC

⇒ B = C

We know that sum of all the angles of a triangle is 180 degrees.

⇒ A+B+C=180

⇒ A+B+B=180

⇒ A+2B=180

Divide both sides by 2

$\frac{A}{2}+B=90$

$\implies\:tan(\frac{A}{2}+B)=tan90$

$\implies\:\frac{tan\frac{A}{2}+tanB}{1-tan\frac{A}{2}\:tanB}=∞$

$\implies\:1-tan\frac{A}{2}\:tanB=0$

$\implies\:1=tan\frac{A}{2}\:tanB$

$\implies\:\frac{1}{tanB}=tan\frac{A}{2}$

$\implies\:cotB=tan\frac{A}{2}$

$\implies\:tan\frac{A}{2}=cotB$

Answer 2

Answer:

Step-by-step explanation:

A+B+C=180

BCOZ Isosceles B=C

A+2B=180

Divide by 2 for A/2

A/2+B=90

A/2=90-B

Multiply both side with tan

tan A/2 = tan(90-B)

tan A/2= cot B