If cosec theta is equal to …

We have,
 {tex}cosec \;A = \frac { \text { Hypotenuse } } { \text { Perpendicular } } = \frac { \sqrt { 10 } } { 1 }{/tex}
So, we draw a right triangle ABC, right-angled at B such that
Perpendicular = BC = 1 unit and, Hypotenuse{tex}= A C = \sqrt { 10 }{/tex}.
By Pythagoras theorem, we have
{tex}A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }{/tex}
{tex}\Rightarrow \quad ( \sqrt { 10 } ) ^ { 2 } = A B ^ { 2 } + 1 ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = 10 - 1 = 9{/tex}
{tex}\Rightarrow \quad A B = \sqrt { 9 } = 3{/tex}
When we consider the trigonometric ratios of {tex}\angle A{/tex}, we have
Base = AB = 3 units, Perpendicular = BC = 1 units and, Hypotenuse{tex}= A C = \sqrt { 10 }{/tex} units
{tex}\therefore \quad \sin A = \frac { \text { Perpendicular } } { \text { Hypotenuse } } = \frac { 1 } { \sqrt { 10 } } , \cos A = \frac { \text { Base } } { \text { Hypotenuse } } = \frac { 3 } { \sqrt { 10 } }{/tex}
 {tex}\tan A = \frac { \text { Perpendicular } } { \text { Base } } = \frac { 1 } { 3 } , \quad \sec A = \frac { \text { Hypotenuse } } { \text { Base } } = \frac { \sqrt { 10 } } { 3 }{/tex}.
and, {tex}\cot A = \frac { \text { Base } } { \text { Perpendicular } } = \frac { 3 } { 1 } = 3{/tex}