If A, B and C are the interior angles of $\Delta ABC$, then what is the value of $\{\tan \frac{A}{2}+\mathrm{cosec}\frac{B+C}{2}\}\{\tan \frac{A}{2}-\mathrm{cosec}\frac{B+C}{2}\}.$

The correct option is –1.

Let us solve this problem step-by-step.

Since A, B, and C are the interior angles of a triangle $\Delta ABC$, we know that the sum of the angles is 180 degrees:
$A+B+C=180°$

From this relationship, we can express the sum $B+C$ in terms of $A$:
$B+C=180°-A$

Dividing both sides by 2 gives:
$\frac{B+C}{2}=90°-\frac{A}{2}$

Now, let us evaluate the term $\mathrm{cosec}\frac{B+C}{2}$ using the co-function trigonometric identity:
$\mathrm{cosec}\frac{B+C}{2}=\mathrm{cosec}(90°-\frac{A}{2})=\sec \frac{A}{2}$

Substitute this back into the expression we want to evaluate:
$\{\tan \frac{A}{2}+\mathrm{cosec}\frac{B+C}{2}\}\{\tan \frac{A}{2}-\mathrm{cosec}\frac{B+C}{2}\}$
This simplifies to:
$\{\tan \frac{A}{2}+\sec \frac{A}{2}\}\{\tan \frac{A}{2}-\sec \frac{A}{2}\}$

Using the algebraic identity $(x+y)(x-y)=x^2-y^2$, we get:
${\tan }^2\frac{A}{2}-{\sec }^2\frac{A}{2}$

Recall the standard trigonometric identity:
${\sec }^2\theta -{\tan }^2\theta =1$
Therefore, it follows that:
${\tan }^2\theta -{\sec }^2\theta =-1$

By substituting $\theta =\frac{A}{2}$, we get:
${\tan }^2\frac{A}{2}-{\sec }^2\frac{A}{2}=-1$

Thus, the value of the given expression is –1.