Question

for triangle ABC prove that TAN (A+C/2)=COT B/2

Answer 1

Answer:

$tan\big(\frac{\angle A+ \angle C}{2}\big)=cot(\frac{B}{2})$

Step-by-step explanation:

$We \: know \: that,\\</p><p>In \: Triangle \: ABC ,</p><p>\\\angle A + \angle B + \angle C=180\degree$

( Angle sum property )

$\implies \angle A+\angle C = 180\degree - \angle B$

Divide both sides by 2 , we get

$\implies \frac{\angle A+ \angle C}{2}=\frac{180}{2}-\frac{B}{2}$

$\implies \frac{\angle A+ \angle C}{2}=90\degree -\frac{B}{2}$

$\implies tan\big(\frac{\angle A+ \angle C}{2}\big)=tan\big(90\degree -\frac{B}{2}\big)$

$\implies tan\big(\frac{\angle A+ \angle C}{2}\big)= cot(\frac{B}{2})$

/* Since ,

tan(90° - A) =cotA */

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Answer 2

Step-by-step explanation:

Let Triangle ABC be a right triangle, right angled at B.

So, Angle B= 90°, so, angle A + angle C=90° ----(1)

(If angle B is 90° so sum of the other two angles should be 90°, so that, angle A+ angle B+angle C=180°)

From (1), angle B= angle a + angle C. -----(2)

Therefore, tan(C+A/2)=cot(C+A/2) [From (1)]

(B can be written as C+A because both are 90°)

Or, tan(C+A/2)=cot(B/2) [From (2)].