SOLUTION: Show that (a) (sec^4-1)/(tan^2x) =2tan^2x (b) (cosec^4x -1)/(cot^2x) =2+cot^2x (c) (1+sec@)/(sin@+tan@)=cosec@ (d) (cos^3@-sin^2@)/(cos@-sin@)=1+sin@cos@ (e) (1-(sin@-cos@)^

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Question 1140527: Show that
(a) (sec^4-1)/(tan^2x) =2tan^2x
(b) (cosec^4x -1)/(cot^2x) =2+cot^2x
(c) (1+sec@)/(sin@+tan@)=cosec@
(d) (cos^3@-sin^2@)/(cos@-sin@)=1+sin@cos@
(e) (1-(sin@-cos@)^2/sin@=2cos@

Found 2 solutions by Edwin McCravy, MathTherapy:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
The limit is two problems.  I'll do a) and d), the ones
which are NOT identities because they were copied wrong:

(a) (sec^4-1)/(tan^2x) =2tan^2x 

The x was left off in the first term and the right side should be tan^4x



Factor the numerator as the difference of squares 



Then we use the identity  rewritten as
this: 







Use the identity   again:





------------------------------------

For (d)   

(d) (cos^3@-sin^2@)/(cos@-sin@)=1+sin@cos@

the sin^2@ should be sin^3@.

(d) 

The top on the left side is the difference of cubes, and
you should remember from algebra that the difference of
cubes factors according to this rule: 

So the numerator factors as

(d) 

   

  

  

Now use the identity  to make
the left side equal to the right side.

Edwin

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Show that
(a) (sec^4-1)/(tan^2x) =2tan^2x
(b) (cosec^4x -1)/(cot^2x) =2+cot^2x
(c) (1+sec@)/(sin@+tan@)=cosec@
(d) (cos^3@-sin^2@)/(cos@-sin@)=1+sin@cos@
(e) (1-(sin@-cos@)^2/sin@=2cos@
(a) should be: 
We'll focus on the left-side:
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