# Find the other dive trigonometric functions OF the acute angle A, given that sec A=2 ?

Apr 15, 2018

Given,secA = 2.
So, cosA = 1/2 = cos 60.
Then, A = 60 degree.
Now, Sin 60 = $\frac{\sqrt{3}}{2}$,tan 60 = $\sqrt{3}$, csc 60 = $\frac{2}{\sqrt{3}}$, cot 60 =$\frac{1}{\sqrt{3}}$.

ANOTHER WAY
We know, secA =$\frac{2}{1} = \frac{h y p o t e \nu s e}{b a s e}$
So, height = $\sqrt{{\left(h y p o t e \nu s e\right)}^{2} - {\left(b a s e\right)}^{2}}$
So, height = $\sqrt{{2}^{2} - {1}^{2}}$
$\Rightarrow \sqrt{3}$

Apr 15, 2018

Given,secA = 2.
So, cosA = 1/2 = cos 60.
Then, A = 60 degree.
Now, Sin 60 = $\frac{\sqrt{3}}{2}$,tan 60 = $\sqrt{3}$, csc 60 = $\frac{2}{\sqrt{3}}$, cot 60 =$\frac{1}{\sqrt{3}}$.

ANOTHER WAY
We know, secA =$\frac{2}{1} = \frac{h y p o t e \nu s e}{b a s e}$
So, height = $\sqrt{{\left(h y p o t e \nu s e\right)}^{2} - {\left(b a s e\right)}^{2}}$
So, height = $\sqrt{{2}^{2} - {1}^{2}}$
$\Rightarrow \sqrt{3}$
Now sinA =$\frac{h e i g h t}{h y p o t e \nu s e} = \frac{\sqrt{3}}{2}$
CosA= $\frac{b a s e}{h y p o t e \nu s e} = \frac{1}{2}$,
TanA = $\frac{h e i g h t}{b a s e} = \frac{\sqrt{3}}{1}$
CscA = 1/sinA =$\frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$,
CotA= 1/tanA =$\frac{1}{\sqrt{3}}$