Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
$L.H.S.\,=\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}$
$=\frac{\cos ^{2} A+(1+\sin A)^{2}}{(1+\sin A)(\cos A)}$
$=\frac{\cos ^{2} A+1+\sin ^{2} A+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{\sin ^{2} A+\cos ^{2} A+1+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{1+1+2 \sin A}{(1+\sin A)(\cos A)}=\frac{2+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{2(1+\sin A)}{(1+\sin A)(\cos A)}=\frac{2}{\cos A}=2 \sec A$
$=R . H . S .$
Express the ratios $\cos A ,$ tan $A$ and $\sec A$ in terms of $\sin A .$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A,$ using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
If $\angle B$ and $\angle Q$ are acute angles such that $\sin B =\sin Q$, then prove that $\angle B =\angle Q$.
$(\sec A+\tan A)(1-\sin A)=..........$
In $Fig.$ find $\tan P-\cot R .$