In $\triangle ABC$,with usual notations,$m \angle C = \frac{\pi}{2}$. If $\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are the roots of the equation $a_1 x^2 + b_1 x + c_1 = 0$ $(a_1 \neq 0)$,then:

The sum of the sides of a right-angled triangle is $42$,and the difference between the median and altitude drawn from the vertex at the right angle is $2$. The area of the triangle is

If in a triangle $ABC$,$a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$,then its sides will be in

In a triangle $ABC$,$BC=7$,$AC=8$,$AB=\alpha \in N$ and $\cos A=\frac{2}{3}$. If $49 \cos (3C)+42=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to..........

In $\triangle ABC$,if $a=2, b=\sqrt{6}$,and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

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 $\cos R > \frac{p}{q}$