Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$L.H.S. =\frac{1+\sec A }{\sec A }=\frac{1+\frac{1}{\cos A }}{\frac{1}{\cos A }}$
$=\frac{\frac{\cos A+1}{\cos A}{1}}{\frac{1}{\cos A}}=(\cos A+1)$
$=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}$
$=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A}$
$= R.H.S.$
State whether the following are true or false. Justify your answer.
$(i)$ $\cos A$ is the abbreviation used for the cosecant of angle $A$
$(ii)$ cot $A$ is the product of cot and $A$.
$(iii)$ $\sin \theta=\frac{4}{3}$ for some angle $\theta$.
In $\triangle ABC ,$ right-angled at $B , AB =24 \,cm , BC =7 \,cm .$ Determine:
$(i)$ $\sin A, \cos A$
$(ii)$ $\sin C, \cos C$
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
Evaluate the following:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
If $\angle B$ and $\angle Q$ are acute angles such that $\sin B =\sin Q$, then prove that $\angle B =\angle Q$.