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 $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$
Using the identity $\operatorname{cosec}^{2} A =1+\cot ^{2} A$
$L.H.S.$ $=\frac{\cos A -\sin A +1}{\cos A +\sin A -1}$
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}+\frac{1}{\sin A}}$
$=\frac{\cot A-1+\operatorname{cosec} A}{\cot A+1-\operatorname{cosec} A}$
$=\frac{\{(\cot A)-(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}{\{(\cot A)+(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}$
$=\frac{(\cot A-1+\operatorname{cosec} A)^{2}}{(\cot A)^{2}-(1-\operatorname{cosec} A)^{2}}$
$=\frac{\cot ^{2} A+1+\operatorname{cosec}^{2} A-2 \cot A-2 \operatorname{cosec} A+2 \cot A \operatorname{cosec} A}{\cot ^{2} A-\left(1+\operatorname{cosec}^{2} A-2 \operatorname{cosec} A\right)}$
$=\frac{2 \operatorname{cosec}^{2} A+2 \cot A \operatorname{cosec} A-2 \cot A-2 \operatorname{cosec} A}{\cot ^{2} A-1-\operatorname{cosec}^{2} A+2 \operatorname{cosec} A}$
$=\frac{2 \operatorname{cosec} A(\operatorname{cosec} A+\cot A)-2(\cot A+\operatorname{cosec} A)}{\cot ^{2} A-\operatorname{cosec}^{2} A-1+2 \operatorname{cosec} A}$
$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{-1-1+2 \operatorname{cosec} A}$
$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{(2 \operatorname{cosec} A-2)}$
$=\operatorname{cosec} A+\cot A$
$= R . H.S.$
Express the trigonometric ratios $\sin A , \sec A$ and $\tan A$ in terms of $\cot A$.
State whether the following are true or false. Justify your answer.
$\sin \theta=\cos \theta$ for all values of $\theta$
If $\tan 2 A=\cot \left(A-18^{\circ}\right),$ where $2 A$ is an acute angle, find the value of $A .$ (in $^{\circ}$)
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$