Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$L.H.S. =\frac{1+\sec A }{\sec A }=\frac{1+\frac{1}{\cos A }}{\frac{1}{\cos A }}$
$=\frac{\frac{\cos A+1}{\cos A}{1}}{\frac{1}{\cos A}}=(\cos A+1)$
$=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}$
$=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A}$
$= R.H.S.$
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
In a right triangle $A B C$, right-angled at $B$. if $\tan A =1,$ then verify that $2 \sin A \cos A=1$
Prove that $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$
State whether the following are true or false. Justify your answer.
$\cot$ $A$ is not defined for $A =0^{\circ}$
Express the trigonometric ratios $\sin A , \sec A$ and $\tan A$ in terms of $\cot A$.