R K Bansal Solutions for Exercise 1: EXERCISE

To prove $\sqrt{\frac{1-\cos \mathrm{A}}{1+\cos \mathrm{A}}}=\frac{\sin \mathrm{A}}{1+ \cos \mathrm{A}}$

Solve the following $\sqrt{\frac{1- \sin \mathrm{A}}{1+ \sin \mathrm{A}}} =\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}$

Solve the following: $1-\frac{{\cos }^2 \mathrm{A}}{1+ \sin A}=\sin \mathrm{A}$

Solve the following $\frac{1}{\sin \mathrm{A}+\cos \mathrm{A}}+\frac{1}{\sin {\displaystyle }{\displaystyle \mathrm{A}}{\displaystyle -}{\displaystyle \cos }{\displaystyle }{\displaystyle \mathrm{A}}}=\frac{2 \sin \mathrm{A}}{1-2 \cos { }^2 \mathrm{A}}$

Solve the following $\frac{\sin \mathrm{A}+\cos \mathrm{A}}{\sin \mathrm{A}-\cos \mathrm{A}}+\frac{\sin {\displaystyle }{\displaystyle \mathrm{A}}{\displaystyle -}{\displaystyle \cos }{\displaystyle }{\displaystyle \mathrm{A}}}{\sin {\displaystyle }{\displaystyle \mathrm{A}}{\displaystyle +}{\displaystyle \cos }{\displaystyle }{\displaystyle \mathrm{A}}}=\frac{2}{2{\sin }^2 \mathrm{A} -1}$

Solve the following $\frac{\cot \mathrm{A}+\mathrm{cosec} \mathrm{A} -1}{\cot \mathrm{A}-\mathrm{cosec} \mathrm{A} +1}=\frac{1+ \cos \mathrm{A}}{\sin {\displaystyle }{\displaystyle \mathrm{A}}}$

Prove the following:

$\frac{\sin \mathrm{\theta } \tan \mathrm{\theta }}{1-\cos \mathrm{\theta }}=1+ \sec \mathrm{\theta }$

Solve the following $\frac{\cos \mathrm{\theta } \cot \mathrm{\theta }}{1+ \sin \mathrm{\theta }}= \mathrm{cosec} \mathrm{\theta }-1$