prove the identity cot(x)-tan(x)/ cot(x) + 1 = 1-tan(x)?

Question

Prove the identity cotx-tanx/cotx+1 = 1-tanx?
I am SO stuck! i have tried a lot of ways and cannot figure out how to prove this....

cot(x) - tan (x) / cot(x) +1 = 1- tan(x)

please help! thanks so much!

Answer 1

cot(x) = cos(x)/sin(x) and tan(x) = sin(x)/cos(x)



So, you're equation says: cot(x) - tan (x) / cot(x) +1 = 1- tan(x) //Let's break this down.



We just stated what cot(x) is:

Now, if we do tan(x)/cot(x) , we basically have this [sin(x)/cos(x)]/[cos(x)/sin(x)] .. After flipping the bottom, and using the dividing/reciporcal rule, we get that this is (sin^2(x)/cos^2(x))



Now, the official equation looks like this:



cos(x)/sin(x) - (sin^2(x)/cos^2(x)) +1 = 1-tan(x) ..... Cancel out the same terms on the right sude and you have:



- sin(x)/cos(x) +1 .... and since we know that sin(x)/cos(x) = tan(x) ... we have:



-tan(x) + 1 .... or 1-tan(x)





Hope this helps. If you need me to write it out on a board, ill do so. u can e-mail me at:

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Answer 2

Maybe i will be able to get you started my solutions are as a rule for algebra its simply that possibly I see something (1+tanx) / (1+cotx) = (1-tanx) / (cotx-1) the right denominator 1/(cotx-1)=1/(-1+cotx)= -1/(1-cotx) so (1-tanx)/(cotx-1)= -(1-tanx)/(1-cotx) offers you (1+tanx) / (1+cotx)= -(1-tanx)/(1-cotx) now you can make the denominators into 'change of squares' left aspect (1+tanx)/(1+cotx)*(1-cotx)/(1-cotx)=(1... Correct side -(1-tanx)/(1-cotx)*(1+cotx)/(1+cotx)= -(1-tanx)(1+cotx)/(1-cotx^2) the (1-cotx^2)'s cancel leaving (1+tanx)(1-cotx)= -(1-tanx)(1+cotx) possibly you can take it from there? Just right good fortune

Answer 3

1/tanx - tanx

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

1/tanx + 1



1 - tan²x

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

1 + tanx



(1 - tanx)(1 + tanx)

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

1 + tanx



1 - tanx