Question

How to you simplify cotx+tanx?

Answer 1

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)`

Answer 2

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)`