In a $\triangle ABC$,if $\angle C = 90^{\circ}$ and $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,then which of the following is true?

If $2 \cos \theta + \sin \theta = 1$,then the value of $4 \cos \theta + 3 \sin \theta$ is equal to

$ABCD$ is a rhombus. The circumradii of $\Delta ABD$ and $\Delta ACD$ are $\frac{25}{2}$ and $25$ respectively. Then the area of the rhombus is .............. $sq. \, unit$.

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

Let $PQR$ be a triangle of area $\Delta$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2}$,where $a, b$ and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2P}$ equals

In $\triangle ABC$,if $B=90^{\circ}$,then $2(r+R)=$