How do you prove: #secx - cosx = sinx tanx#?
4 Answers
Using the definitions of
First convert all terms into
Second apply fraction sum rules to the LHS.
Lastly we apply the Pythagorean identity:
Explanation:
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.
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#
It's really this simple...
Explanation:
Using the identity
Then, multiply
Considering that
Finally, using the trigonometric identity