How would you prove $tanx/cscx = secx - 1/secx$ ?

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Answer 1 · 2016-05-07T19:00:49

Replace all simple functions
$tan x/csc x = tanx sin x = sin^2x/cos x$
$1/cosx - 1/secx = 1/cosx - cos x = sin^2x/cosx$

Answer 2 · 2016-05-09T02:08:55

Just to clarify the answer of the previous contributor.

On the left side, we must apply the identities $tanx = sinx/cosx$ and $cscx= 1/sinx$

$(sinx/cosx)/(1/sinx) = 1/cosx - 1/(1/cosx)$

$(sinx xx sinx)/cosx = 1/cosx - cosx$

$sin^2x/cosx = (1 - cos^2x)/cosx$

Applying the Pythagorean identity $1 - cos^2x = sin^2x$

$sin^2x/cosx = sin^2x/cosx$

Identity proved!!

Hopefully this helps!

How would you prove tanx/cscx = secx - 1/secxtanx/cscx = secx - 1/secx?