(tan\theta -cot\theta )/(tan\theta +cot\theta )+2cos^(2)\theta =1

Question

(tanΘ -cotΘ )/(tanΘ +cotΘ )   +  2cos²Θ =1

Answer 1

Hi Derek.

 

The key here is to convert both the tan(theta) and cot(theta) to sines and cosines.

 

The problem becomes:  (I am skipping the theta's for now - makes typing faster.  will put the back in before it is all over...)

 

sin      cos

---   -  ---

cos      sin

-------------   + 2 cos²(theta) = 1

sin      cos

---  +  ---

cos      sin

 

Turn this complex fraction into a single fraction - get a common denominator for the top, and a common denominator for the bottom.

 

 

sin²   -   cos²

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

   sin  cos

-----------------  + 2 cos²(theta) = 1

sin²  +  cos²

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

    sin  cos

 

 

Now, multiply the fraction on top and bottom by  (sin  cos) and you will no longer have a fraction of fractions, but a single fraction that looks like

 

sin² - cos²

--------------   + 2 cos²(theta) = 1

sin² + cos²    

 

 

Remember the trig identity:  sin²(theta) + cos²(theta) = 1, so the denominator of the fraction = 1.

 

Now what you have is  

 

sin²(theta) - cos²(theta) + 2cos²(theta) = 1

 

The two "cos²" terms are like terms and can be combined, so the left side is now

 

sin²(theta) + cos²(theta) = 1

 

and the same identity as above makes the left side 1.  So

 

 

1 = 1  and it is verified.

 

 :-)