Question

How do you verify `(tanx) /(1-cotx) + (cotx) /(1-tanx) = 1 +secx cscx`?

Answer 1

Carefully.

Explanation:

I find that its easier to convert directly to sine and cosine. I recognize the trig identities better.

`(sinX/cosX)/(1-cosX/sinX) + (cosX/sinX)/(1-sinX/cosX)`

There are too many fractions. Turn the `1`'s into `sinX"/"sinX` and `cosX"/"cosX`, then combine the denominators into fractions over `sinX` and `cosX`.

`(sinX/cosX)/((sinX-cosX)/sinX) + (cosX/sinX)/((cosX-sinX)/cosX)`

Now we can get rid of these fractions of fractions by flipping the denominators and multiplying them by the numerators.

`sin^2X/(cosX(sinX-cosX)) + cos^2X/(sinX(cosX-sinX))`

Cross multiply the denominators to get a common denominator.

`(sin^3X(cosX-sinX) + cos^3(sinX-cosX))/(sinXcosX(sinX-cosX)(cosX-sinX))`

Multiply through the parenthesis.

`(sin^3XcosX-sin^4X - cos^4X+sinXcos^3X)/(sinXcosX(-sin^2X+2sinXcosX - cos^2X))`

Pull a factor of `sinXcosX` out of two of the terms in the numerator. There is also a `-(sin^2X+cos^2X)` in the denominator, which by the Pythagorean theorem is equal to `-1`.

`((sin^2X + cos^2X)sinXcosX -sin^4X - cos^4X)/(sinXcosX(2sinXcosX-1))`

We can use the Pythagorean theorem again on the top. Also, pull out a `-1` from the `""^4` terms in the numerator.

`(sinXcosX -(sin^4X + cos^4X))/(sinXcosX(2sinXcosX-1))`

Split the `sin^4X` term into two `sin^2X` terms. Then we can use the Pythagorean theorem, `sin^2X = 1-cos^2X` to replace one of the `sin^2` terms. Do a similar action with the `cosX^4` term.

`(sinXcosX -(sin^2Xsin^2X+cos^2Xcos^2X))/(sinXcosX(2sinXcosX-1))`

`(sinXcosX -(sin^2X(1-cos^2X)+cos^2X(1-sin^2X)))/(sinXcosX(2sinXcosX-1))`

Multiply through the terms in the top parenthesis.

`(sinXcosX -(sin^2X-2sin^2Xcos^2X +cos^2X))/(sinXcosX(2sinXcosX-1))`

Once again Pythagorean theorem out the `sin^2X` and `cos^2X` terms.

`(sinXcosX -(1-2sin^2Xcos^2X))/(sinXcosX(2sinXcosX-1))`

Multiply the `-1` through the parenthesis in the numerator.

`( sinXcosX+2sin^2Xcos^2X-1)/(sinXcosX(2sinXcosX-1))`

Pull out a factor of `sinXcosX` from two of the terms in the numerator.

`( sinXcosX(1+2sinXcosX)-1)/(sinXcosX(2sinXcosX-1))`

Adding and subtracting a `1` inside the parenthesis is the same as adding `0`.

`( sinXcosX(1+2sinXcosX+(1-1))-1)/(sinXcosX(2sinXcosX-1))`

Combine the two `+1` terms in the parenthesis.

`( sinXcosX(2sinXcosX-1 +2)-1)/(sinXcosX(2sinXcosX-1))`

Now pull the `2` out. Remember to multiply by `sinXcosX`.

`( sinXcosX(2sinXcosX-1) + 2sinXcosX-1)/(sinXcosX(2sinXcosX-1))`

Finally we have a term in the numerator that is the same as the denominator. Split the addition terms in the numerator to get;

`(sinXcosX(2sinXcosX-1))/(sinXcosX(2sinXcosX-1)) + (2sinXcosX-1)/(sinXcosX(2sinXcosX-1))`

The first term simplifies to `1` and in the second term the `(2sinXcosX-1)`s cancel out.

`1+1/(sinXcosX)`

Now convert to `secX`and `cscX`

`1+secXcscX`

Answer 2

`LHS=tanx/(1 - cotx) + cotx/(1 - tanx)`

`=tan^2x/(tanx(1 - cotx)) + cotx/(1 - tanx)`

`=cotx/(1 - tanx) +tan^2x/(tanx- tanxcotx)`

`=cotx/(1 - tanx) -tan^2x/(1- tanx)`

`=(1/tanx-tan^2x)/(1 - tanx)`

`=(1-tan^3x)/(tanx(1 - tanx))`

`=(cancel((1-tanx))(1+tanx+tan^2x))/(tanxcancel((1 - tanx))`

`=(tanx+sec^2x)/(tanx`

`=tanx/tanx+sec^2x/tanx`

`=1+secx*secx *(cosx/sinx)`

`=1+secx*(1/sinx)`

`=1+secx*cscx=RHS`