If $\sin ( A - B )=\frac{1}{2}, \cos ( A + B )=\frac{1}{2}, 0^{\circ} < A + B \leq 90^{\circ}, A > B ,$ find $A$ and $B$
since, $\sin ( A - B )=\frac{1}{2},$ therefore, $A - B =30^{\circ}$ ......$(1)$
Also, since $\cos ( A + B )=\frac{1}{2},$ therefore, $A + B =60^{\circ}$ ......$(2)$
Solving $(1)$ and $(2),$ we get $: A=45^{\circ}$ and $B=15^{\circ} .$
Evaluate:
$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$
Evaluate the following:
$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$
If $\tan ( A + B )=\sqrt{3}$ and $\tan ( A - B )=\frac{1}{\sqrt{3}} ; 0^{\circ}< A + B \leq 90^{\circ} ; A > B ,$ find $A$ and $B$
Evaluate:
$\cos 48^{\circ}-\sin 42^{\circ}$