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Simplify \(\cot 10 + \tan 5.\)

Enter your answer as a trigonometric function evaluated at an integer, such as "sin 7".

 Apr 1, 2020
 #4
avatar+20876 
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To write  cot(10) + tan(5) as one trig function:

 

First, to make this both more general and also easier to type, let's have  x = 5  and  2x = 10.

 

Now, the problem is:  cot(2x) + tan(x)

 

But,  cot(2x)  =  1/tan(2x)  

 

Since  tan(2x)  =  2tan(x) / [ 1 - tan2(x) ],  so   cot(2x)  =  [1 - tan2(x)] / [2tan(x)]

 

This means that we have:  [1 - tan2(x)] / [2tan(x)] + tan(x)

 

Writing both with the common denominator of  2tan(x):

--->   [1 - tan2(x)] / [2tan(x)] + 2tan2(x) / [2tan(x)]

--->   [1 + tan2(x)] / [2tan(x)]

 

Since  1 + tan2(x)  =  sec2(x)

--->   sec2(x) / [2tan(x)]

 

sec2(x)  =  1/cos2(x)     and     tan(x)  =  sin(x)/cos(x)

--->   [1/cos2(x)] / [2·sin(x)/cos(x)]  =  cos(x) / [2·sin(x)·cos2(x)]  =  1 / [2sin(x)cos(x)

 

But  sin(x)cos(x)  =  ½sin(2x)

--->   1 / [2·½·sin(2x)  =  1 / sin(2x)  =  csc(2x)

 

Since  x = 5   --->   1 / sin(10) =  csc(10)

 Apr 2, 2020

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