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

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

We should write

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

Now, we use the Pythagorean identity, which is

#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

#1/cosx-cosx=secx-cosx#

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Using the definitions of

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First convert all terms into

Second apply fraction sum rules to the LHS.

Lastly we apply the Pythagorean identity:

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First in questions of these forms it's a good idea to convert all terms into sine and cosine: so, replace

and replace

The LHS,

The RHS,

Now we apply fraction sum rules to the LHS, making a common base (just like number fraction like

LHS=

Lastly we apply the Pythagorean identity:

By rearranging it we get

We replace the

LHS =

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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It's really this simple...

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Using the identity

Then, multiply

Considering that

Finally, using the trigonometric identity

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