Assam Board Solutions for Chapter: Introduction to Trigonometry, Exercise 4: Exercise 8.4

Prove the following identity, where the angles involved are acute angles for which the expressions is defined.
${\left(\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)}^2=\frac{1-\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}$
 

Prove the following identity where the angle involved is acute angle for which the expression is defined.
$\frac{1+\mathrm{secA}}{\mathrm{secA}}=\frac{{\sin }^2\mathrm{A}}{1-\mathrm{cosA}}$

Prove the following identity, where the angles involved are acute angles for which the expression is defined.

$\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}=\mathrm{cosec}\mathrm{A}+\cot \mathrm{A}$

Prove the following identity, where the angle involved is acute angle for which the expression is defined.
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$
 

Prove the following identity where the angle involved is acute angle for which the following expression is defined.
$\frac{\sin \mathrm{\theta }-2{\sin }^3\mathrm{\theta }}{2{\cos }^3\mathrm{\theta }-\cos \mathrm{\theta }}=\tan \mathrm{\theta }$
 

Prove the following identity, where the angles involved are acute angles for which the expression is defined.
${\left(\sin \mathrm{A}+\mathrm{cosec}\mathrm{A}\right)}^2+{\left(\cos \mathrm{A}+\sec \mathrm{A}\right)}^2=7+{\tan }^2\mathrm{A}+{\cot }^2\mathrm{A}$
 

Prove the following identity where the angle involved is acute angle for which the expression is defined.
$\left(\mathrm{cosec}\mathrm{A}-\sin \mathrm{A}\right)\left(\sec \mathrm{A}-\cos \mathrm{A}\right)=\frac{1}{\tan \mathrm{A}+\cot \mathrm{A}}$
 

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\left(\frac{1+{\tan }^2\mathrm{A}}{1+{\cot }^2\mathrm{A}}\right)={\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)}^2={\tan }^2\mathrm{A}$