If in a $\Delta ABC$,$\cos A \cos B + \sin A \sin B \sin^2 C = 1$,then the statement which is incorrect is:

If the reciprocals of the lengths of the sides of a $\triangle ABC$ are in harmonic progression,then its ex-radii $r_1, r_2, r_3$ are in

If in $\triangle ABC$,with usual notations,$a^2, b^2, c^2$ are in $A$.$P$.,then $\frac{\sin 3B}{\sin B} =$

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$,respectively. Then which of the following statements is (are) $TRUE$?
$(A)$ $\cos P \geq 1-\frac{p^2}{2qr}$
$(B)$ $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
$(C)$ $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$
$(D)$ If $p < q$ and $p < r$,then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$

If the area of a triangle $ABC$ is $\Delta$, then ${a^2}\sin 2B + {b^2}\sin 2A$ is equal to (in $\Delta$)

If in $\Delta ABC,$ $a = 6, b = 3$ and $\cos(A - B) = \frac{4}{5},$ then its area will be ..... $square \, unit.$