Question 1140527
<pre>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

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

Factor the numerator as the difference of squares {{{A^2-B^2=(A-B)(A+B)}}}

{{{((sec^2(x)^""-1)(sec^2(x)^""+1))/(tan^2(x)) =tan^4(x)}}}

Then we use the identity {{{1+tan^2(theta)=sec^2(theta)}}} rewritten as
this: {{{sec^2(theta)-1=tan^2(theta)}}}

{{{(tan^2(x)^""(sec^2(x)^""-1))/(tan^2(x)^"") =tan^4(x)}}}

{{{(cross(tan^2(x)^"")(sec^2(x)^""-1))/(cross(tan^2(x)^"")) =tan^4(x)}}}

{{{tan^2(x)^""(sec^2(x)^""-1) =tan^4(x)}}}

Use the identity  {{{sec^2(theta)-1=tan^2(theta)}}} again:

{{{tan^2(x)tan^2(x) =tan^4(x)}}}

{{{tan^4(x)=tan^4(x)}}}

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

For (d)   

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

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

(d) {{{(cos^3(theta)-sin^3(theta))/(cos(theta)-sin(theta))=1+sin(theta)cos(theta)}}}

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: {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}

So the numerator factors as

(d) {{{(cos(theta)-sin(theta)^"")(cos^2(theta)+cos(theta)sin(theta)+sin^2(theta)^"")/(cos(theta)-sin(theta))=1+sin(theta)cos(theta)^""}}}

   {{{(cross(cos(theta)-sin(theta)^""))(cos^2(theta)+cos(theta)sin(theta)+sin^2(theta)^"")/(cross(cos(theta)^""-sin(theta)))=1+sin(theta)cos(theta)}}}

  {{{(cross(cos(theta)-sin(theta)^""))(cos^2(theta)+cos(theta)sin(theta)+sin^2(theta)^"")/(cross(cos(theta)^""-sin(theta)))=1+sin(theta)cos(theta)}}}

  {{{cos^2(theta)+cos(theta)sin(theta)+sin^2(theta)=1+sin(theta)cos(theta)^""}}}

Now use the identity {{{cos^2(theta)+sin^2(theta)=1}}} to make
the left side equal to the right side.

Edwin</pre>