Question

If abc are interior angles of triangle abc the find the value of tan a/2*tan b+c/2

Answer 1

Solution

The value of tan(A/2) × tan(B + C)/2 is 1

Given

A,B and C are the interior angles of ∆ABC

To finD

tan(A/2) × tan(B + C)/2

In ∆ABC,

A + B + C = 180°. [Angle Sum Property]

→ B + C = 180 - A

Dividing both sides by 2,

→ (B + C)/2 = 90 - A/2

Multiplying both sides by tan,

→ tan(B + C)/2 = tan[90 - A/2]

→ tan(B + C)/2 = cot A/2

Now,

tan(A/2) × tan(B+ C)/2

= tan(A/2) cot(A/2)

= 1

Answer 2

$\bf Given:\ \tt A,\ B\ and\ C\ are\ the\ interior\ angles\ of\ \triangle\ ABC.\\\\\\\bf To\ find:\ \tt The\ value\ of\ tan\ \dfrac{A}{2}\ \times\ tan\ \dfrac{B\ +\ C}{2}.\\\\\\\bf Answer:\\\\\\\tt We\ know\ that\ the\ interior\ angles\ in\ a\ triangle\ add\ upto\ {180}^{\circ}.\\\\\\Therefore,\ \\\\\\A\ +\ B\ +\ C\ =\ {180}^{\circ}.\\\\\\\implies\ B\ +\ C\ =\ {180}^{\circ}\ -\ A.\\\\\\\bigg[Dividing\ both\ sides\ by\ 2\bigg],$

$\Longrightarrow \tt \ \ \dfrac{B\ +\ C}{2}\ =\ \dfrac{{180}^{\circ}\ -\ A}{2}\\\\\\\implies\ \dfrac{B\ +\ C}{2}\ =\ {90}^{\circ}\ -\ \dfrac{A}{2}\\\\\\\bigg[Multiplying\ both\ sides\ with\ tan\bigg],\\\\\\\implies\ tan\ \bigg(\dfrac{B\ +\ C}{2}\bigg)\ =\ tan\ \bigg({90}^{\circ}\ -\ \dfrac{A}{2}\bigg)\\ \\\\\bigg[tan\ ({90}^{\circ}\ -\ \theta)\ =\ cot\ \theta\bigg]\\\\\\\implies\ tan\ \bigg(\dfrac{B\ +\ C}{2}\bigg)\ =\ cot\ \bigg(\dfrac{A}{2}\bigg)$

$\tt Now,\ we\ know\ that\ tan\ \bigg(\dfrac{B\ +\ C}{2}\bigg)\ =\ cot\ \bigg(\dfrac{A}{2}\bigg).\\\\\\According\ to\ the\ question,\ we\ need\ to\ find\ tan\ \dfrac{A}{2}\ \times\ tan\ \dfrac{B\ +\ C}{2}.\\\\\\\implies\ tan\ \dfrac{A}{2}\ \times\ cot\ \dfrac{A}{2}\\\\\\\bigg[tan\ \theta\ =\ \dfrac{1}{cot\ \theta}\bigg]\\\\\\\implies\ \dfrac{1}{cot\ \frac{A}{2}}\ \times\ cot\ \dfrac{A}{2}\\\\\\=\ \ \ \ \bf 1$