How do you prove $csc x = sec (\frac{π}{2} - x)$ $cscx = sec(pi/2 - x)$ ?

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How do you prove $csc x = sec (\frac{π}{2} - x)$ $cscx = sec(pi/2 - x)$ ?

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Answer 1 · 2018-03-29T03:25:09

Since this is an identity,

the relation is true for all values of x

Explanation:

Given:

$csc x = sec (\frac{π}{2} - x)$ $cscx=sec(pi/2-x)$

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

$sec (\frac{π}{2} - x) = \frac{1}{cos (\frac{π}{2} - x)}$ $sec(pi/2-x)=1/cos(pi/2-x)$

Thus,

$\frac{1}{sin x} = \frac{1}{cos (\frac{π}{2} - x)}$ $1/sinx=1/cos(pi/2-x)$

$cos (\frac{π}{2} - x) = sin x$ $cos(pi/2-x)=sinx$

Since this is an identity,

the relation is true for all values of x

Answer 2 · 2018-03-29T03:25:56

See below

Explanation:

Using:
$sec x = \frac{1}{cos x}$ $secx=1/cosx$
$\frac{1}{sin x} = csc x$ $1/sinx=cscx$
$cos (x - y) = cos x cos y + sin x sin y$ $cos(x-y)=cosxcosy+sinxsiny$

Start:
$csc x = sec (\frac{π}{2} - x)$ $cscx=sec(pi/2-x)$

$csc x = \frac{1}{cos (\frac{π}{2} - x)}$ $cscx=1/cos(pi/2-x)$

$csc x = \frac{1}{cos (\frac{π}{2}) cos x + sin (\frac{π}{2}) \cdot sin x}$ $cscx=1/(cos(pi/2)cosx+sin(pi/2)*sinx)$

$cscx=1/(cancel(0*cosx)+1*sinx)$

$cscx=1/sinx$

$cscx=cscx$

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How do you prove $csc x = sec (\frac{π}{2} - x)$ $cscx = sec(pi/2 - x)$ ?

2 Answers

Explanation:

Explanation:

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