How to you simplify cotx+tanx?

2 Answers
Narad T.
Dec 7, 2017

The answer is #=2csc(2x)#

Explanation:

We need

#tanx=sinx/cosx#

#cotx=cosx/sinx#

#cos^2x+sin^2s=1#

#sin2x=2sinxcosx#

#cscx=1/sinx#

Therefore,

#cotx+tanx=cosx/sinx+sinx/cosx#

#=(cos^2x+sin^2x)/(sinxcosx)#

#=1/(sinxcosx)#

#=2/sin(2x)#

#=2csc(2x)#

Alvin L.
Dec 7, 2017

Alternatively, we can get #sec(x)csc(x)#

Explanation:

If we write #cot(x)# as #1/tan(x)#, we get:
#cot(x)+tan(x)=1/tan(x)+tan(x)#

Then we bring under a common denominator:
#=1/tan(x)+(tan(x)*tan(x))/tan(x)=(1+tan^2(x))/tan(x)#

Now we can use the #tan^2(x)+1=sec^2(x)# identity:
#=sec^2(x)/tan(x)#

To try and work out some of the relationships between these functions, let's represent the functions in terms of a right triangle. We'll let #a# be the adjacent, #b# be the opposite and #c# be the hypotenuse. This gives:
#=((c/a)^2)/(b/a)=c^2/a^2*a/b=c^2/(ab)=c/a*c/b#

Now if we convert back to trigonometric functions, we get:
#c/a*c/b=sec(x)csc(x)#

This is equal to Narad T's answer of #2csc(2x)#

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