How do you prove: secx - cosx = sinx tanxsecxcosx=sinxtanx?

4 Answers
Feb 11, 2016

Using the definitions of secxsecx and tanxtanx, along with the identity
sin^2x + cos^2x = 1sin2x+cos2x=1, we have

secx-cosx = 1/cosx-cosxsecxcosx=1cosxcosx

=1/cosx-cos^2x/cosx=1cosxcos2xcosx

=(1-cos^2x)/cosx=1cos2xcosx

=sin^2x/cosx=sin2xcosx

=sinx *sinx/cosx=sinxsinxcosx

=sinxtanx=sinxtanx

Feb 11, 2016

First convert all terms into sinxsinx and cosxcosx.
Second apply fraction sum rules to the LHS.
Lastly we apply the Pythagorean identity: sin^2 x + cos^2 x =1 sin2x+cos2x=1

Explanation:

First in questions of these forms it's a good idea to convert all terms into sine and cosine: so, replace tan xtanx with sin x /cos xsinxcosx
and replace sec x secx with 1/ cos x1cosx.

The LHS, sec x- cos xsecxcosx becomes 1/cos x- cos x1cosxcosx.
The RHS, sin x tan xsinxtanx becomes sin x sin x/cos x sinxsinxcosx or sin^2 x / cos xsin2xcosx.

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)13+14412+312=712).
LHS=1/cos x- cos x => 1/cos x- cos^2 x/cos x => {1 - cos^2 x} /cos x1cosxcosx1cosxcos2xcosx1cos2xcosx.

Lastly we apply the Pythagorean identity: sin^2 x + cos^2 x =1 sin2x+cos2x=1! (one of the most useful identities for these types of problems).
By rearranging it we get 1- cos^2 x = sin^2 x1cos2x=sin2x.
We replace the 1- cos^2 x 1cos2x in the LHS with sin^2 xsin2x.

LHS = {1 - cos^2 x} /cos x => {sin^2 x} /cos x1cos2xcosxsin2xcosx 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.

Sep 7, 2016

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

We should write sinxtanxsinxtanx in terms of sinxsinx and cosxcosx, using the identity color(red)(tanx=sinx/cosx)tanx=sinxcosx:

sinxtanx=sinx(sinx/cosx)=sin^2x/cosxsinxtanx=sinx(sinxcosx)=sin2xcosx

Now, we use the Pythagorean identity, which is sin^2x+cos^2x=1sin2x+cos2x=1. We can modify this to solve for sin^2xsin2x, so: color(red)(sin^2x=1-cos^2x)sin2x=1cos2x:

sin^2x/cosx=(1-cos^2x)/cosxsin2xcosx=1cos2xcosx

Now, just split up the numerator:

(1-cos^2x)/cosx=1/cosx-cos^2x/cosx=1/cosx-cosx1cos2xcosx=1cosxcos2xcosx=1cosxcosx

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

1/cosx-cosx=secx-cosx1cosxcosx=secxcosx

Sep 8, 2016

It's really this simple...

Explanation:

Using the identity tanx=sinx/cosxtanx=sinxcosx, multiply the sinxsinx onto the identity to get:

secx-cosx=sin^2x/cosxsecxcosx=sin2xcosx

Then, multiply cosxcosx through the equation to yield:

1-cos^2x=sin^2x1cos2x=sin2x

Considering that secxsecx is the inverse of cosxcosx.
Finally, using the trigonometric identity 1-cos^2x=sin^2x1cos2x=sin2x, the final answer would be:

sin^2x=sin^2xsin2x=sin2x