How do you simplify #(sec x - cos x) / tan x#?

Question

Zoe Austin · Accepted Answer

#sinx#

This can also be proven by showing that #secx-cosx = (tanx)(sinx)# and then dividing both sides by #tanx# Let's start by breaking down some terms. In my opinion, you have to kind of play around with trig stuff to get it to break down right. #secx=1/cosx=tanx/sinx# So, #(secx-cosx)/tanx = secx/tanx - cosx/tanx = (tanx/sinx)/tanx - cosx/tanx# #=1/sinx - cosx/tanx# Tangent = sine/cosine, so the reciprocal of the tangent = cosine/sine #= 1/sinx - cos^2x/sinx = (1-cos^2x)/sinx# Since #sin^2x+cos^2x=1#, that means #cos^2x=1-sin^2x# #= (1-(1-sin^2x))/sinx = (1 - 1 + sin^2x)/sinx = sin^2x/sinx = sinx# Final Answer

Madelyn Adams · Answer

undefined To simplify $ \frac{\sec(x) - \cos(x)}{\tan(x)} $, we can use trigonometric identities.

Recall that $ \sec(x) = \frac{1}{\cos(x)} $ and $ \tan(x) = \frac{\sin(x)}{\cos(x)} $.

Substitute these identities into the expression:

$$ \frac{\frac{1}{\cos(x)} - \cos(x)}{\frac{\sin(x)}{\cos(x)}} $$

Combine the fractions:

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

Now, recall the Pythagorean identity: $ 1 - \cos^2(x) = \sin^2(x) $.

Substitute this into the expression:

$$ = \frac{\sin^2(x)}{\sin(x)} $$

Simplify:

$$ = \sin(x) $$

So, $ \frac{\sec(x) - \cos(x)}{\tan(x)} $ simplifies to $ \sin(x) $.