How do you prove: #secx - cosx = sinx tanx#?

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mason m Share
Sep 7, 2016

If we so desire, we can also modify the right-hand side to match the left-hand side.

We should write #sinxtanx# in terms of #sinx# and #cosx#, using the identity #color(red)(tanx=sinx/cosx)#:

#sinxtanx=sinx(sinx/cosx)=sin^2x/cosx#

Now, we use the Pythagorean identity, which is #sin^2x+cos^2x=1#. We can modify this to solve for #sin^2x#, so: #color(red)(sin^2x=1-cos^2x)#:

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

Now, just split up the numerator:

#(1-cos^2x)/cosx=1/cosx-cos^2x/cosx=1/cosx-cosx#

Use the reciprocal identity #color(red)(secx=1/cosx#:

#1/cosx-cosx=secx-cosx#

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sente Share
Feb 11, 2016

Using the definitions of #secx# and #tanx#, along with the identity
#sin^2x + cos^2x = 1#, we have

#secx-cosx = 1/cosx-cosx#

#=1/cosx-cos^2x/cosx#

#=(1-cos^2x)/cosx#

#=sin^2x/cosx#

#=sinx *sinx/cosx#

#=sinxtanx#

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Maggie Share
Feb 11, 2016

Answer:

First convert all terms into #sinx# and #cosx#.
Second apply fraction sum rules to the LHS.
Lastly we apply the Pythagorean identity: # sin^2 x + cos^2 x =1 #

Explanation:

First in questions of these forms it's a good idea to convert all terms into sine and cosine: so, replace #tan x# with #sin x /cos x#
and replace #sec x # with #1/ cos x#.

The LHS, #sec x- cos x# becomes #1/cos x- cos x#.
The RHS, # sin x tan x# becomes #sin x sin x/cos x # or #sin^2 x / cos x#.

Now we apply fraction sum rules to the LHS, making a common base (just like number fraction like #1/3 +1/4 => 4/12 + 3/12 = 7/12)#.
LHS=#1/cos x- cos x => 1/cos x- cos^2 x/cos x => {1 - cos^2 x} /cos x#.

Lastly we apply the Pythagorean identity: # sin^2 x + cos^2 x =1 #! (one of the most useful identities for these types of problems).
By rearranging it we get #1- cos^2 x = sin^2 x#.
We replace the #1- cos^2 x # in the LHS with #sin^2 x#.

LHS = # {1 - cos^2 x} /cos x => {sin^2 x} /cos x# which is equal to the modified RHS.

Thus LHS= RHS Q.E.D.

Note this general pattern of getting things into terms of sine and cosine, using the fraction rules and the Pythagorean identity, often solves these types of questions.

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Sep 8, 2016

Answer:

It's really this simple...

Explanation:

Using the identity #tanx=sinx/cosx#, multiply the #sinx# onto the identity to get:

#secx-cosx=sin^2x/cosx#

Then, multiply #cosx# through the equation to yield:

#1-cos^2x=sin^2x#

Considering that #secx# is the inverse of #cosx#.
Finally, using the trigonometric identity #1-cos^2x=sin^2x#, the final answer would be:

#sin^2x=sin^2x#

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