Question

express sec theta in all trigonometric ratios

Answer 1

I hope this will help you

Answer 2

Result 1
$\sec(x) = \frac{1}{ \cos(x) }$

Now,As we know that
Result 2
$\sec^{2} (x) - \tan^{2} (x) = 1$

$\sec(x) = \sqrt{1 - \tan^{2} (x) }$

Now, Result 3

$\sec(x) = \frac{1}{ \cos(x) }$

and as we know that

$\sin^{2} (x ) + \cos^{2} (x) = 1$

So

$\sec(x) = \frac{1}{ \sqrt{1 - \sin^{2} (x) } }$

Now, Result 4

as we know that

$\sin(x) = \frac{1}{ \cosec(x) }$
so, In result 3

$\sec(x) = \frac{1}{ \sqrt{1 - \sin^{2} (x) } }$

$\sec(x) = \frac{1}{ \sqrt{1 - \frac{1}{ \cosec^{2} ( {x} ) } } }$

$\sec(x) = \frac{1}{ \sqrt{ \frac{ \cosec^{2} (x) - 1}{ \cosec^{2} ( {x} ) } } }$

$\sec(x) = \frac{ \sqrt{ \cosec ^{2} (x) } }{ \sqrt{ { \cosec^{2} (x) - 1} } }$

$\sec(x) = \frac{ { \cosec (x) } }{ \sqrt{ { \cosec^{2} (x) - 1} } }$
Now, Result 5 from result 2

$\sec(x) = \sqrt{1 - \tan^{2} (x) }$

$\sec(x) = \sqrt{1 - \frac{1}{\cot^{2} (x) } }$

$\sec(x) = \sqrt{ \frac{ \cot^{2} (x) - 1}{ \cot^{2} (x ) } }$

$\sec(x) = \frac{ \sqrt{ \cot^{2} (x) - 1} }{ \cot(x) }$
Done.