Melanie Z.

asked • 03/18/16

(sinx+cosxcotx)/(cosxcscx)=secx

Trig Identities 

Inactive Tutor

I hate to nitpick on the other tutors' solutions, but I see something that really isn't good form. The two other solutions show one side of the identity carried down for one or more steps. The really proper way to show and identity is to start with one side at a time and change it (using algebra and other trigonometric identities) step-by-step to equal the other side.
 
Again, I am nitpicking about this, but there is a reason why I am. The process of solving one of these problems is not just to get it right, but to train the student to think ahead without already knowing the answer. Knowing the outcome is helpful, but in more advanced math courses, YOU NEVER KNOW WHAT THE FINAL OUTCOME WILL BE. Trigonometric identities are fabulous tools for simplifying many things. In calculus, especially, you do not always know what you are simplifying to, just that you need to do some simplifying to make a problem easier. Sometimes, the final result does NOT look simpler than the original, but still allows you to make progress on a problem where before nothing else could be done to solve it.
 
I really have no complaint about the actual MATH used in the other tutors' answers, just that their approaches, if followed routinely, could cause you problems later on. Please keep this in mind when doing other problems like this. Having taught this subject many times for almost three decades, I think I know a little about what can happen when the form is slightly different from what the teacher intended. It may seem pointless, but I assure you that it is not.
 
Even if you have already decided not to take any math beyond this course, you should not shortchange yourself on learning a good method which can train your mind to solve similar problems in other subjects. Then again, you might find that learning to do this my way may actually make you WANT to study more math in the future.
 
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03/18/16

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Inactive Tutor answered • 03/18/16

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Hi Melanie, I have two different ways to do it.  Here is 1st one:
 
(sin x+cos x cot x)/(cos x csc x) = sec x
 
Rewrite:
 
(sin x+cos x cot x)
--------------------         =    sec x
  (cos x csc x)
 
split into 2 fractions:
 
(sin x)                 (cos x cot x)
-------------   +        --------------        = sec x      [cos x cancels out]
(cos x csc x)       (cos x csc x)
 
 
(sin x) (sin x)           cot x 
-----------------  +       -------   = sec x                  cot x = cos x csc x
     (cos x)               csc x   
 
(sin x) (sin x)           (cos x) (csc x)
--------------      +        ---------------   =   sec x        [csc x cancels out]
      cos x                      csc x
 
sin2x           cos x
-------     +   -------  =  sec x                            
 cos x             1
 
sin2x           cos2x
------      +    -------   = sec x
cos x           cos x
 
 
sin2x + cos2x                                   1
---------------   =   sec x                   ------- = sec x
       cos x                                      cos x
 
 
 
 
 
 
Here is 2nd way:
 
(sin x+cos x cot x)/(cos x csc x) = sec x
 
cot x = cos x csc x
 
 
Rewrite
 
(sin x+cos x cot x) / cot x = sec x
 
sin x       cos x cot x
-----  +     ------------  =  sec x
cot x          cot x
 
sin x          cos x
-----     +     -------  =  sec x
cot x            1
 
 
cot x = cos x / sin x
 
 
(sin x) (sin x)         cos x
--------------     +       ------   =   sec x
     cos x                  1
 
sin2x         cos2x                                  1
------     +    -------   =   sec x             -------- = sec x
cos x         cos x                                cos x
 
 
I hope you understood.  Good luck to you.
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Arthur D. answered • 03/18/16

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(sinx+cosxcotx)/(cosxcscx)=secx
cotx=cosx/sinx
cscx=1/sinx
(sinx+cosx[cosx/sinx])/(cosx[1/sinx])=secx
(sinx+cos^2x/sinx)/(cosx/sinx)=secx
([sin^2x+cos^2x]/sinx)÷(cosx/sinx)=secx
(1/sinx)÷(cosx/sinx)=seccx
(1/sinx)(sinx/cosx)=secx
(1/cosx)=secx
secx=secx
 
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Melanie Z.

I dont understand how you jump from (sinx+cos^2x/sinx)/(cosx/sinx)=secx to
([sin^2x+cos^2x]/sinx)÷(cosx/sinx)=secx 
could you please explain that 
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03/19/16

Arthur D.

tutor
sinx + cos^2x/sinx
 
write sinx as sin^2x/sinx so the two fractions have a common denominator
 
sin^2x/sinx + cos^2x/sinx
 
add the fractions
 
(sin^2x+cos^2x)/sinx
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03/19/16

Inactive Tutor answered • 03/18/16

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Start with the more complicated side and rewrite in terms of sines and cosines:
 
(sin x + cos x cot x)/(cos x csc x) = (sin x + cos x · cos x/sin x)/(cosx · 1/sin x)
     = (sin2 x + cos2 x)/sin x · sin x/ cos x
     = 1/sin x · sin x/cos x
     = 1/cos x
     = sec x.
 
Voila! The second line was found by getting a common denominator for the first set of parentheses and then inverting the denominator and multiplying by the reciprocal. The rest should be pretty easy to follow.
 
Post a reply if you have any questions.
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Arthur D.

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Alan,
      My solution is identical to your solution. The student can already see the right side and if you can make the left side, for example, equal the right side, then you have proven the identity. Also, sometimes it's easier to work with one side at a time and show that both sides equal  the same thing. I really don't understand your concern about my solution or Yohan's solution.
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03/19/16

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