Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We just started this today in class and I have several homework problems, but I don't understand it at all. Can anyone show me step by step how to do this. The problem is $4/(\tan x + \cot x)$.

My teacher went ahead and told us the answer would be $4\sin x\cos x$, so we can make sure we use the right steps and get the correct answer. I am so lost in this. Please help me understand.

share|improve this question
    
Please elaborate on what to do this this expression. What are the instructions? –  King Squirrel May 8 at 21:50
    
My teacher just said to show the steps using the trig identities to get to the answer.I guess it is considered solving it? Not sure. Sorry –  Ila Isabelle May 8 at 21:53

2 Answers 2

up vote 1 down vote accepted

Assuming that your teacher wants the expression simplified:

4/(tanx+cotx)

4/((sin/cosx)+(cosx/sinx))

4/((sin^2x/sinxcosx)+(cos^2x/sinxcosx))

4/((sin^2x+cos^2x)/(sinxcosx))

//sin^2x+cos^2x=1

=> 4/(1/(sinxcosx))

4sinxcosx

share|improve this answer
    
Thank you very much! –  Ila Isabelle May 8 at 21:58
    
The King orders Ila Isabelle to choose his answer by clicking the check mark. And the King says your welcome. –  King Squirrel May 8 at 22:00
1  
Lol, as you command King. –  Ila Isabelle May 8 at 22:03

Use the definition: $$ \tan x=\frac{\sin x}{\cos x},\quad \cot x=\frac{\cos x}{\sin x}. $$ So $$ \frac{4}{\tan x+\cot x}=\frac{4}{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}}= \frac{4}{\dfrac{\sin^2 x+\cos^2x}{\sin x\cos x}} $$ Now, what can you do?

share|improve this answer
    
sin^2x + cos^2x= 1, so then I multiply 4*sinx*cosx. Please tell me I am right? –  Ila Isabelle May 8 at 21:56
    
@IlaIsabelle Yes, you are. –  egreg May 8 at 21:57
    
Thank you so much!! –  Ila Isabelle May 8 at 21:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.