Question

If $$\sec { \theta } =2$$, then find the values of other t-ratios of angle $$\theta$$.

Answer 1

We know that
$$\sec { \theta } =\cfrac { hypotenuse }{ base } $$
$$\sec { \theta } =\cfrac { 2 }{ 1 } \Rightarrow \cfrac { H }{ B } =\cfrac { 2 }{ 1 } \Rightarrow \cfrac { AC }{ BC } =\cfrac { 2 }{ 1 } $$
let $$BC=1k$$ and $$AC=2k$$
where $$k$$ is any positive integer
In right angled $$\triangle ABC$$,we have
$${ \left( AB \right) }^{ 2 }+{ \left( 1k \right) }^{ 2 }={ \left( 2k \right) }^{ 2 }\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( k \right) }^{ 2 }={ 4k }^{ 2 }\Rightarrow { \left( AB \right) }^{ 2 }={ 4k }^{ 2 }-{ k }^{ 2 }={ 3k }^{ 2 }\Rightarrow AB=k\sqrt { 3 } $$ (By Pythagoras theorem)
Now, we have to find the value of other trigonometric ratios
we know that
$$\sin { \theta } =\cfrac { Perpendicular }{ hypotenuse } =\cfrac { AB }{ AC } =\cfrac { k\sqrt { 3 } }{ 2k } =\cfrac { \sqrt { 3 } }{ 2 } $$
$$\cos { \theta } =\cfrac { \quad Base }{ hypotenuse } =\cfrac { BC }{ AC } =\cfrac { 1k }{ 2k } =\cfrac { 1 }{ 2 } $$
$$\tan { \theta } =\cfrac { Perpendicular }{ Base } =\cfrac { AB }{ BC } =\cfrac { k\sqrt { 3 } }{ k } =\cfrac { \sqrt { 3 } }{ 1 } =\sqrt { 3 } $$
$$co\sec { \theta } =\cfrac { 1 }{ \sin { \theta } } =\cfrac { 1 }{ \cfrac { \sqrt { 3 } }{ 2 } } =\cfrac { 2 }{ \sqrt { 3 } } ;\cot { \theta } =\cfrac { 1 }{ \tan { \theta } } =\cfrac { 1 }{ \sqrt { 3 } } $$