Question

State whether the following statement is true or false.

If $A,B,C$ are interior angles of $ABC$, then $\text{cosec}\left(\frac{A+B}{2}\right)$ $=\sec$ $\left(\frac{C}{2}\right)$ .

• A. True
• B. False

Answer 1

The correct option is A True

Here, $ABC$ is a triangle and $A,B,C$ are interior angles of the triangle.

$\therefore$ $A+B+C={180}^o$ [ Angle sum property of a triangle ]

$\Rightarrow$ $A+B={180}^o-C$

Dividing both sides by $2$ we get,

$\Rightarrow$ $\frac{A+B}{2}=\frac{{180}^o}{2}-\frac{C}{2}$

$\Rightarrow$ $\frac{A+B}{2}={90}^o-\frac{C}{2}$

Multiplying both sides by $cosec$ we get,

$\Rightarrow$ $cosec\left(\frac{A+B}{2}\right)=cosec({90}^o-\frac{C}{2})$

We know, $cosec({90}^o-\theta )=\sec \theta$

$\Rightarrow$ $cosec\left(\frac{A+B}{2}\right)=\sec \left(\frac{C}{2}\right)$ ------ Hence proved